Limit definition of partial derivative calculator

嚜燉imit definition of partial derivative calculator

4.3.1 Calculate the partial derivatives of a function of two variables. 4.3.2 Calculate the partial derivatives of a function of more than two variables. 4.3.3 Determine the higher-order derivatives of a function of two variables. 4.3.4 Explain the meaning of a partial differential equation and give an example. Now that we have examined limits and

continuity of functions of two variables, we can proceed to study derivatives. Finding derivatives of functions of two variables is the key concept in this chapter, with as many applications in mathematics, science, and engineering as differentiation of single-variable functions. However, we have already seen that limits and continuity of multivariable

functions have new issues and require new terminology and ideas to deal with them. This carries over into differentiation as well. When studying derivatives of functions of one variable, we found that one interpretation of the derivative is an instantaneous rate of change of yy as a function of x.x. Leibniz notation for the derivative is dy/dx,dy/dx, which

implies that yy is the dependent variable and xx is the independent variable. For a function z=f(x,y)z=f(x,y) of two variables, xx and yy are the independent variables and zz is the dependent variable. This raises two questions right away: How do we adapt Leibniz notation for functions of two variables? Also, what is an interpretation of the derivative?

The answer lies in partial derivatives. Let f(x,y)f(x,y) be a function of two variables. Then the partial derivative of ff with respect to x,x, written as ?f/?x,?f/?x, or fx,fx, is defined as ?f?x=limh↙0f(x+h,y)?f(x,y)h.?f?x=limh↙0f(x+h,y)?f(x,y)h. The partial derivative of ff with respect to y,y, written as ?f/?y,?f/?y, or fy,fy, is defined as ?f?y=limk↙0f(x,y+k)

?f(x,y)k.?f?y=limk↙0f(x,y+k)?f(x,y)k. This definition shows two differences already. First, the notation changes, in the sense that we still use a version of Leibniz notation, but the dd in the original notation is replaced with the symbol ?.?. (This rounded ※d§※d§ is usually called ※partial,§ so ?f/?x?f/?x is spoken as the ※partial of ff with respect to x.§)x.§)

This is the first hint that we are dealing with partial derivatives. Second, we now have two different derivatives we can take, since there are two different independent variables. Depending on which variable we choose, we can come up with different partial derivatives altogether, and often do. Use the definition of the partial derivative as a limit to

calculate ?f/?x?f/?x and ?f/?y?f/?y for the function f(x,y)=x2?3xy+2y2?4x+5y?12.f(x,y)=x2?3xy+2y2?4x+5y?12. First, calculate f(x+h,y).f(x+h,y). f ( x + h , y ) = ( x + h ) 2 ? 3 ( x + h ) y + 2 y 2 ? 4 ( x + h ) + 5 y ? 12 = x 2 + 2 x h + h 2 ? 3 x y ? 3 h y + 2 y 2 ? 4 x ? 4 h + 5 y ? 12. f ( x + h , y ) = ( x + h ) 2 ? 3 ( x + h ) y + 2 y 2 ? 4 ( x + h ) + 5

y ? 12 = x 2 + 2 x h + h 2 ? 3 x y ? 3 h y + 2 y 2 ? 4 x ? 4 h + 5 y ? 12. Next, substitute this into Equation 4.12 and simplify: ? f ? x = lim h ↙ 0 f ( x + h , y ) ? f ( x , y ) h = lim h ↙ 0 ( x 2 + 2 x h + h 2 ? 3 x y ? 3 h y + 2 y 2 ? 4 x ? 4 h + 5 y ? 12 ) ? ( x 2 ? 3 x y + 2 y 2 ? 4 x + 5 y ? 12 ) h = lim h ↙ 0 x 2 + 2 x h + h 2 ? 3 x y ? 3 h y + 2 y 2 ? 4 x

? 4 h + 5 y ? 12 ? x 2 + 3 x y ? 2 y 2 + 4 x ? 5 y + 12 h = lim h ↙ 0 2 x h + h 2 ? 3 h y ? 4 h h = lim h ↙ 0 h ( 2 x + h ? 3 y ? 4 ) h = lim h ↙ 0 ( 2 x + h ? 3 y ? 4 ) = 2 x ? 3 y ? 4. ? f ? x = lim h ↙ 0 f ( x + h , y ) ? f ( x , y ) h = lim h ↙ 0 ( x 2 + 2 x h + h 2 ? 3 x y ? 3 h y + 2 y 2 ? 4 x ? 4 h + 5 y ? 12 ) ? ( x 2 ? 3 x y + 2 y 2 ? 4 x + 5 y ? 12 ) h =

lim h ↙ 0 x 2 + 2 x h + h 2 ? 3 x y ? 3 h y + 2 y 2 ? 4 x ? 4 h + 5 y ? 12 ? x 2 + 3 x y ? 2 y 2 + 4 x ? 5 y + 12 h = lim h ↙ 0 2 x h + h 2 ? 3 h y ? 4 h h = lim h ↙ 0 h ( 2 x + h ? 3 y ? 4 ) h = lim h ↙ 0 ( 2 x + h ? 3 y ? 4 ) = 2 x ? 3 y ? 4. To calculate ?f?y,?f?y, first calculate f(x,y+h):f(x,y+h): f ( x , y + h ) = x 2 ? 3 x ( y + h ) + 2 ( y + h ) 2 ? 4 x + 5 (

y + h ) ? 12 = x 2 ? 3 x y ? 3 x h + 2 y 2 + 4 y h + 2 h 2 ? 4 x + 5 y + 5 h ? 12. f ( x , y + h ) = x 2 ? 3 x ( y + h ) + 2 ( y + h ) 2 ? 4 x + 5 ( y + h ) ? 12 = x 2 ? 3 x y ? 3 x h + 2 y 2 + 4 y h + 2 h 2 ? 4 x + 5 y + 5 h ? 12. Next, substitute this into Equation 4.13 and simplify: ? f ? y = lim k ↙ 0 f ( x , y + h ) ? f ( x , y ) k = lim k ↙ 0 ( x 2 ? 3 x y ? 3 x k

+ 2 y 2 + 4 y k + 2 k 2 ? 4 x + 5 y + 5 k ? 12 ) ? ( x 2 ? 3 x y + 2 y 2 ? 4 x + 5 y ? 12 ) k = lim k ↙ 0 x 2 ? 3 x y ? 3 x k + 2 y 2 + 4 y k + 2 k 2 ? 4 x + 5 y + 5 k ? 12 ? x 2 + 3 x y ? 2 y 2 + 4 x ? 5 y + 12 k = lim k ↙ 0 ?3 x k + 4 y k + 2 k 2 + 5 k k = lim k ↙ 0 h ( ?3 x + 4 y + 2 k + 5 ) k = lim k ↙ 0 ( ?3 x + 4 y + 2 k + 5 ) = ?3 x + 4 y + 5. ? f ? y =

lim k ↙ 0 f ( x , y + h ) ? f ( x , y ) k = lim k ↙ 0 ( x 2 ? 3 x y ? 3 x k + 2 y 2 + 4 y k + 2 k 2 ? 4 x + 5 y + 5 k ? 12 ) ? ( x 2 ? 3 x y + 2 y 2 ? 4 x + 5 y ? 12 ) k = lim k ↙ 0 x 2 ? 3 x y ? 3 x k + 2 y 2 + 4 y k + 2 k 2 ? 4 x + 5 y + 5 k ? 12 ? x 2 + 3 x y ? 2 y 2 + 4 x ? 5 y + 12 k = lim k ↙ 0 ?3 x k + 4 y k + 2 k 2 + 5 k k = lim k ↙ 0 h ( ?3 x + 4 y + 2 k +

5 ) k = lim k ↙ 0 ( ?3 x + 4 y + 2 k + 5 ) = ?3 x + 4 y + 5. Use the definition of the partial derivative as a limit to calculate ?f/?x?f/?x and ?f/?y?f/?y for the function f(x,y)=4x2+2xy?y2+3x?2y+5.f(x,y)=4x2+2xy?y2+3x?2y+5. The idea to keep in mind when calculating partial derivatives is to treat all independent variables, other than the variable

with respect to which we are differentiating, as constants. Then proceed to differentiate as with a function of a single variable. To see why this is true, first fix yy and define g(x)=f(x,y)g(x)=f(x,y) as a function of x.x. Then g∩(x)=limh↙0g(x+h)?g(x)h=limh↙0f(x+h,y)?f(x,y)h=?f?x.g∩(x)=limh↙0g(x+h)?g(x)h=limh↙0f(x+h,y)?f(x,y)h=?f?x. The same is

true for calculating the partial derivative of ff with respect to y.y. This time, fix xx and define h(y)=f(x,y)h(y)=f(x,y) as a function of y.y. Then h∩(x)=limk↙0h(x+k)?h(x)k=limk↙0f(x,y+k)?f(x,y)k=?f?y.h∩(x)=limk↙0h(x+k)?h(x)k=limk↙0f(x,y+k)?f(x,y)k=?f?y. All differentiation rules from Introduction to Derivatives apply. Calculate ?f/?x?f/?x and ?f/?y?f/

?y for the following functions by holding the opposite variable constant then differentiating: f(x,y)=x2?3xy+2y2?4x+5y?12f(x,y)=x2?3xy+2y2?4x+5y?12 g(x,y)=sin(x2y?2x+4)g(x,y)=sin(x2y?2x+4) To calculate ?f/?x,?f/?x, treat the variable yy as a constant. Then differentiate f(x,y)f(x,y) with respect to xx using the sum, difference, and power rules:

?f?x=??x[x2?3xy+2y2?4x+5y?12]=??x[x2]???x[3xy]+??x[2y2]???x[4x]+??x[5y]???x[12]=2x?3y+0?4+0?0=2x?3y?4.?f?x=??x[x2?3xy+2y2?4x+5y?12]=??x[x2]???x[3xy]+??x[2y2]???x[4x]+??x[5y]???x[12]=2x?3y+0?4+0?0=2x?3y?4. The derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable x,x, so

they are treated as constant terms. The derivative of the second term is equal to the coefficient of x,x, which is ?3y.?3y. Calculating ?f/?y:?f/?y: ?f?y=??y[x2?3xy+2y2?4x+5y?12]=??y[x2]???y[3xy]+??y[2y2]???y[4x]+??y[5y]???y[12]=?3x+4y?0+5?0=?3x+4y+5.?f?y=??y[x2?3xy+2y2?4x+5y?12]=??y[x2]???y[3xy]+??y[2y2]???y[4x]+??y[5y]

???y[12]=?3x+4y?0+5?0=?3x+4y+5. These are the same answers obtained in Example 4.14. To calculate ?g/?x,?g/?x, treat the variable y as a constant. Then differentiate g(x,y)g(x,y) with respect to x using the chain rule and power rule: ?g?x=??x[sin(x2y?2x+4)]=cos(x2y?2x+4)??x[x2y?2x+4]=

(2xy?2)cos(x2y?2x+4).?g?x=??x[sin(x2y?2x+4)]=cos(x2y?2x+4)??x[x2y?2x+4]=(2xy?2)cos(x2y?2x+4). To calculate ?g/?y,?g/?y, treat the variable xx as a constant. Then differentiate g(x,y)g(x,y) with respect to yy using the chain rule and power rule:

?g?y=??y[sin(x2y?2x+4)]=cos(x2y?2x+4)??y[x2y?2x+4]=x2cos(x2y?2x+4).?g?y=??y[sin(x2y?2x+4)]=cos(x2y?2x+4)??y[x2y?2x+4]=x2cos(x2y?2x+4). Calculate ?f/?x?f/?x and ?f/?y?f/?y for the function f(x,y)=tan(x3?3x2y2+2y4)f(x,y)=tan(x3?3x2y2+2y4) by holding the opposite variable constant, then differentiating. How can we interpret these

partial derivatives? Recall that the graph of a function of two variables is a surface in ?3.?3. If we remove the limit from the definition of the partial derivative with respect to x,x, the difference quotient remains: f(x+h,y)?f(x,y)h.f(x+h,y)?f(x,y)h. This resembles the difference quotient for the derivative of a function of one variable, except for the

presence of the yy variable. Figure 4.21 illustrates a surface described by an arbitrary function z=f(x,y).z=f(x,y). Figure 4.21 Secant line passing through the points ( x , y , f ( x , y ) ) ( x , y , f ( x , y ) ) and ( x + h , y , f ( x + h , y ) ) . ( x + h , y , f ( x + h , y ) ) . In Figure 4.21, the value of hh is positive. If we graph f(x,y)f(x,y) and f(x+h,y)f(x+h,y) for an

arbitrary point (x,y),(x,y), then the slope of the secant line passing through these two points is given by f(x+h,y)?f(x,y)h.f(x+h,y)?f(x,y)h. This line is parallel to the x-axis.x-axis. Therefore, the slope of the secant line represents an average rate of change of the function ff as we travel parallel to the x-axis.x-axis. As hh approaches zero, the slope of the

secant line approaches the slope of the tangent line. If we choose to change yy instead of xx by the same incremental value h,h, then the secant line is parallel to the y-axisy-axis and so is the tangent line. Therefore, ?f/?x?f/?x represents the slope of the tangent line passing through the point (x,y,f(x,y))(x,y,f(x,y)) parallel to the x-axisx-axis and ?f/?y?f/?y

represents the slope of the tangent line passing through the point (x,y,f(x,y))(x,y,f(x,y)) parallel to the y-axis.y-axis. If we wish to find the slope of a tangent line passing through the same point in any other direction, then we need what are called directional derivatives, which we discuss in Directional Derivatives and the Gradient. We now return to the

idea of contour maps, which we introduced in Functions of Several Variables. We can use a contour map to estimate partial derivatives of a function g(x,y).g(x,y). Use a contour map to estimate ?g/?x?g/?x at the point (5,0)(5,0) for the function g(x,y)=9?x2?y2.g(x,y)=9?x2?y2. The following graph represents a contour map for the function

g(x,y)=9?x2?y2.g(x,y)=9?x2?y2. Figure 4.22 Contour map for the function g ( x , y ) = 9 ? x 2 ? y 2 , g ( x , y ) = 9 ? x 2 ? y 2 , using c = 0 , 1 , 2 , c = 0 , 1 , 2 , and 3 3 ( c = 3 ( c = 3 corresponds to the origin). The inner circle on the contour map corresponds to c=2c=2 and the next circle out corresponds to c=1.c=1. The first circle is given by the

equation 2=9?x2?y2;2=9?x2?y2; the second circle is given by the equation 1=9?x2?y2.1=9?x2?y2. The first equation simplifies to x2+y2=5x2+y2=5 and the second equation simplifies to x2+y2=8.x2+y2=8. The x-interceptx-intercept of the first circle is (5,0)(5,0) and the x-interceptx-intercept of the second circle is (22,0).(22,0). We can estimate

the value of ?g/?x?g/?x evaluated at the point (5,0)(5,0) using the slope formula: ? g ? x | ( x , y ) = ( 5 , 0 ) ＞ g ( 5 , 0 ) ? g ( 2 2 , 0 ) 5 ? 2 2 = 2 ? 1 5 ? 2 2 = 1 5 ? 2 2 ＞ ?1.688 . ? g ? x | ( x , y ) = ( 5 , 0 ) ＞ g ( 5 , 0 ) ? g ( 2 2 , 0 ) 5 ? 2 2 = 2 ? 1 5 ? 2 2 = 1 5 ? 2 2 ＞ ?1.688 . To calculate the exact value of ?g/?x?g/?x evaluated at the point (5,0),

(5,0), we start by finding ?g/?x?g/?x using the chain rule. First, we rewrite the function as g(x,y)=9?x2?y2=(9?x2?y2)1/2g(x,y)=9?x2?y2=(9?x2?y2)1/2 and then differentiate with respect to xx while holding yy constant: ? g ? x = 1 2 ( 9 ? x 2 ? y 2 ) ?1 / 2 ( ?2 x ) = ? x 9 ? x 2 ? y 2 . ? g ? x = 1 2 ( 9 ? x 2 ? y 2 ) ?1 / 2 ( ?2 x ) = ? x 9 ? x 2 ? y 2 .

Next, we evaluate this expression using x=5x=5 and y=0:y=0: ? g ? x | ( x , y ) = ( 5 , 0 ) = ? 5 9 ? ( 5 ) 2 ? ( 0 ) 2 = ? 5 4 = ? 5 2 ＞ ?1.118 . ? g ? x | ( x , y ) = ( 5 , 0 ) = ? 5 9 ? ( 5 ) 2 ? ( 0 ) 2 = ? 5 4 = ? 5 2 ＞ ?1.118 . The estimate for the partial derivative corresponds to the slope of the secant line passing through the points (5,0,g(5,0))

(5,0,g(5,0)) and (22,0,g(22,0)).(22,0,g(22,0)). It represents an approximation to the slope of the tangent line to the surface through the point (5,0,g(5,0)),(5,0,g(5,0)), which is parallel to the x-axis.x-axis. Use a contour map to estimate ?f/?y?f/?y at point (0,2)(0,2) for the function f(x,y)=x2?y2.f(x,y)=x2?y2. Compare this with the exact answer. Suppose

we have a function of three variables, such as w=f(x,y,z).w=f(x,y,z). We can calculate partial derivatives of ww with respect to any of the independent variables, simply as extensions of the definitions for partial derivatives of functions of two variables. Let f(x,y,z)f(x,y,z) be a function of three variables. Then, the partial derivative of ff with respect to x,

written as ?f/?x,?f/?x, or fx,fx, is defined to be ?f?x=limh↙0f(x+h,y,z)?f(x,y,z)h.?f?x=limh↙0f(x+h,y,z)?f(x,y,z)h. The partial derivative of ff with respect to y,y, written as ?f/?y,?f/?y, or fy,fy, is defined to be ?f?y=limk↙0f(x,y+k,z)?f(x,y,z)k.?f?y=limk↙0f(x,y+k,z)?f(x,y,z)k. The partial derivative of ff with respect to z,z, written as ?f/?z,?f/?z, or fz,fz, is

defined to be ?f?z=limm↙0f(x,y,z+m)?f(x,y,z)m.?f?z=limm↙0f(x,y,z+m)?f(x,y,z)m. We can calculate a partial derivative of a function of three variables using the same idea we used for a function of two variables. For example, if we have a function ff of x,y,andz,x,y,andz, and we wish to calculate ?f/?x,?f/?x, then we treat the other two independent

variables as if they are constants, then differentiate with respect to x.x. Use the limit definition of partial derivatives to calculate ?f/?x?f/?x for the function f(x,y,z)=x2?3xy+2y2?4xz+5yz2?12x+4y?3z.f(x,y,z)=x2?3xy+2y2?4xz+5yz2?12x+4y?3z. Then, find ?f/?y?f/?y and ?f/?z?f/?z by setting the other two variables constant and differentiating

accordingly. We first calculate ?f/?x?f/?x using Equation 4.14, then we calculate the other two partial derivatives by holding the remaining variables constant. To use the equation to find ?f/?x,?f/?x, we first need to calculate f(x+h,y,z):f(x+h,y,z): f ( x + h , y , z ) = ( x + h ) 2 ? 3 ( x + h ) y + 2 y 2 ? 4 ( x + h ) z + 5 y z 2 ? 12 ( x + h ) + 4 y ? 3 z = x 2 +

2 x h + h 2 ? 3 x y ? 3 x h + 2 y 2 ? 4 x z ? 4 h z + 5 y z 2 ? 12 x ? 12 h + 4 y ? 3 z f ( x + h , y , z ) = ( x + h ) 2 ? 3 ( x + h ) y + 2 y 2 ? 4 ( x + h ) z + 5 y z 2 ? 12 ( x + h ) + 4 y ? 3 z = x 2 + 2 x h + h 2 ? 3 x y ? 3 x h + 2 y 2 ? 4 x z ? 4 h z + 5 y z 2 ? 12 x ? 12 h + 4 y ? 3 z and recall that

f(x,y,z)=x2?3xy+2y2?4zx+5yz2?12x+4y?3z.f(x,y,z)=x2?3xy+2y2?4zx+5yz2?12x+4y?3z. Next, we substitute these two expressions into the equation: ? f ? x = lim h ↙ 0 [ x 2 + 2 x h + h 2 ? 3 x y ? 3 h y + 2 y 2 ? 4 x z ? 4 h z + 5 y z 2 ? 12 x ? 12 h + 4 y ? 3 z h ? x 2 ? 3 x y + 2 y 2 ? 4 x z + 5 y z 2 ? 12 x + 4 y ? 3 z h ] = lim h ↙ 0 [ 2 x h + h 2

? 3 h y ? 4 h z ? 12 h h ] = lim h ↙ 0 [ h ( 2 x + h ? 3 y ? 4 z ? 12 ) h ] = lim h ↙ 0 ( 2 x + h ? 3 y ? 4 z ? 12 ) = 2 x ? 3 y ? 4 z ? 12. ? f ? x = lim h ↙ 0 [ x 2 + 2 x h + h 2 ? 3 x y ? 3 h y + 2 y 2 ? 4 x z ? 4 h z + 5 y z 2 ? 12 x ? 12 h + 4 y ? 3 z h ? x 2 ? 3 x y + 2 y 2 ? 4 x z + 5 y z 2 ? 12 x + 4 y ? 3 z h ] = lim h ↙ 0 [ 2 x h + h 2 ? 3 h y ? 4 h z ?

12 h h ] = lim h ↙ 0 [ h ( 2 x + h ? 3 y ? 4 z ? 12 ) h ] = lim h ↙ 0 ( 2 x + h ? 3 y ? 4 z ? 12 ) = 2 x ? 3 y ? 4 z ? 12. Then we find ?f/?y?f/?y by holding xandzxandz constant. Therefore, any term that does not include the variable yy is constant, and its derivative is zero. We can apply the sum, difference, and power rules for functions of one variable: ?

? y [ x 2 ? 3 x y + 2 y 2 ? 4 x z + 5 y z 2 ? 12 x + 4 y ? 3 z ] = ? ? y [ x 2 ] ? ? ? y [ 3 x y ] + ? ? y [ 2 y 2 ] ? ? ? y [ 4 x z ] + ? ? y [ 5 y z 2 ] ? ? ? y [ 12 x ] + ? ? y [ 4 y ] ? ? ? y [ 3 z ] = 0 ? 3 x + 4 y ? 0 + 5 z 2 ? 0 + 4 ? 0 = ?3 x + 4 y + 5 z 2 + 4. ? ? y [ x 2 ? 3 x y + 2 y 2 ? 4 x z + 5 y z 2 ? 12 x + 4 y ? 3 z ] = ? ? y [ x 2 ] ? ? ? y [ 3 x y ] + ? ? y [ 2 y 2 ]

? ? ? y [ 4 x z ] + ? ? y [ 5 y z 2 ] ? ? ? y [ 12 x ] + ? ? y [ 4 y ] ? ? ? y [ 3 z ] = 0 ? 3 x + 4 y ? 0 + 5 z 2 ? 0 + 4 ? 0 = ?3 x + 4 y + 5 z 2 + 4. To calculate ?f/?z,?f/?z, we hold x and y constant and apply the sum, difference, and power rules for functions of one variable: ? ? z [ x 2 ? 3 x y + 2 y 2 ? 4 x z + 5 y z 2 ? 12 x + 4 y ? 3 z ] = ? ? z [ x 2 ] ? ? ? z [ 3

x y ] + ? ? z [ 2 y 2 ] ? ? ? z [ 4 x z ] + ? ? z [ 5 y z 2 ] ? ? ? z [ 12 x ] + ? ? z [ 4 y ] ? ? ? z [ 3 z ] = 0 ? 0 + 0 ? 4 x + 10 y z ? 0 + 0 ? 3 = ?4 x + 10 y z ? 3. ? ? z [ x 2 ? 3 x y + 2 y 2 ? 4 x z + 5 y z 2 ? 12 x + 4 y ? 3 z ] = ? ? z [ x 2 ] ? ? ? z [ 3 x y ] + ? ? z [ 2 y 2 ] ? ? ? z [ 4 x z ] + ? ? z [ 5 y z 2 ] ? ? ? z [ 12 x ] + ? ? z [ 4 y ] ? ? ? z [ 3 z ] = 0 ? 0 + 0 ? 4

x + 10 y z ? 0 + 0 ? 3 = ?4 x + 10 y z ? 3. Use the limit definition of partial derivatives to calculate ?f/?x?f/?x for the function f(x,y,z)=2x2?4x2y+2y2+5xz2?6x+3z?8.f(x,y,z)=2x2?4x2y+2y2+5xz2?6x+3z?8. Then find ?f/?y?f/?y and ?f/?z?f/?z by setting the other two variables constant and differentiating accordingly. Calculate the three partial

derivatives of the following functions. f(x,y,z)=x2y?4xz+y2x?3yzf(x,y,z)=x2y?4xz+y2x?3yz g(x,y,z)=sin(x2y?z)+cos(x2?yz)g(x,y,z)=sin(x2y?z)+cos(x2?yz) In each case, treat all variables as constants except the one whose partial derivative you are calculating. ?f?x=??x[x2y?4xz+y2x?3yz]=??x(x2y?4xz+y2)(x?3yz)?(x2y?4xz+y2)??x(x?3yz)

(x?3yz)2=(2xy?4z)(x?3yz)?(x2y?4xz+y2)(1)(x?3yz)2=2x2y?6xy2z?4xz+12yz2?x2y+4xz?y2(x?3yz)2=x2y?6xy2z?4xz+12yz2+4xz?y2(x?3yz)2?f?x=??x[x2y?4xz+y2x?3yz]=??x(x2y?4xz+y2)(x?3yz)?(x2y?4xz+y2)??x(x?3yz)(x?3yz)2=(2xy?4z)(x?3yz)?(x2y?4xz+y2)(1)

(x?3yz)2=2x2y?6xy2z?4xz+12yz2?x2y+4xz?y2(x?3yz)2=x2y?6xy2z?4xz+12yz2+4xz?y2(x?3yz)2 ?f?y=??y[x2y?4xz+y2x?3yz]=??y(x2y?4xz+y2)(x?3yz)?(x2y?4xz+y2)??y(x?3yz)(x?3yz)2=(x2+2y)(x?3yz)?(x2y?4xz+y2)(?3z)

(x?3yz)2=x3?3x2yz+2xy?6y2z+3x2yz?12xz2+3y2z(x?3yz)2=x3+2xy?3y2z?12xz2(x?3yz)2?f?y=??y[x2y?4xz+y2x?3yz]=??y(x2y?4xz+y2)(x?3yz)?(x2y?4xz+y2)??y(x?3yz)(x?3yz)2=(x2+2y)(x?3yz)?(x2y?4xz+y2)(?3z)(x?3yz)2=x3?3x2yz+2xy?6y2z+3x2yz?12xz2+3y2z(x?3yz)2=x3+2xy?3y2z?12xz2(x?3yz)2

?f?z=??z[x2y?4xz+y2x?3yz]=??z(x2y?4xz+y2)(x?3yz)?(x2y?4xz+y2)??z(x?3yz)(x?3yz)2=(?4x)(x?3yz)?(x2y?4xz+y2)(?3y)(x?3yz)2=?4x2+12xyz+3x2y2?12xyz+3y3(x?3yz)2=?4x2+3x2y2+3y3(x?3yz)2?f?z=??z[x2y?4xz+y2x?3yz]=??z(x2y?4xz+y2)(x?3yz)?(x2y?4xz+y2)??z(x?3yz)(x?3yz)2=(?4x)(x?3yz)?(x2y?4xz+y2)(?3y)

(x?3yz)2=?4x2+12xyz+3x2y2?12xyz+3y3(x?3yz)2=?4x2+3x2y2+3y3(x?3yz)2 ?f?x=??x[sin(x2y?z)+cos(x2?yz)]=(cos(x2y?z))??x(x2y?z)?(sin(x2?yz))??x(x2?yz)=2xycos(x2y?z)?2xsin(x2?yz)?f?y=??y[sin(x2y?z)+cos(x2?yz)]=(cos(x2y?z))??y(x2y?z)?(sin(x2?yz))??y(x2?yz)=x2cos(x2y?z)+zsin(x2?yz)?f?z=??z[sin(x2y?z)+cos(x2?yz)]=

(cos(x2y?z))??z(x2y?z)?(sin(x2?yz))??z(x2?yz)=?cos(x2y?z)+ysin(x2?yz)?f?x=??x[sin(x2y?z)+cos(x2?yz)]=(cos(x2y?z))??x(x2y?z)?(sin(x2?yz))??x(x2?yz)=2xycos(x2y?z)?2xsin(x2?yz)?f?y=??y[sin(x2y?z)+cos(x2?yz)]=(cos(x2y?z))??y(x2y?z)?(sin(x2?yz))??y(x2?yz)=x2cos(x2y?z)+zsin(x2?yz)?f?z=??z[sin(x2y?z)+cos(x2?yz)]=

(cos(x2y?z))??z(x2y?z)?(sin(x2?yz))??z(x2?yz)=?cos(x2y?z)+ysin(x2?yz) Calculate ?f/?x,?f/?x, ?f/?y,?f/?y, and ?f/?z?f/?z for the function f(x,y,z)=sec(x2y)?tan(x3yz2).f(x,y,z)=sec(x2y)?tan(x3yz2). Consider the function f(x,y)=2x3?4xy2+5y3?6xy+5x?4y+12.f(x,y)=2x3?4xy2+5y3?6xy+5x?4y+12. Its partial derivatives are

?f?x=6x2?4y2?6y+5and?f?y=?8xy+15y2?6x?4.?f?x=6x2?4y2?6y+5and?f?y=?8xy+15y2?6x?4. Each of these partial derivatives is a function of two variables, so we can calculate partial derivatives of these functions. Just as with derivatives of single-variable functions, we can call these second-order derivatives, third-order derivatives, and so on.

In general, they are referred to as higher-order partial derivatives. There are four second-order partial derivatives for any function (provided they all exist): ?2f?x2=??x[?f?x], ?2f?x?y= ??x[?f?y], ?2f?y?x=??y[?f?x],?2f?y2=??y[?f?y].?2f?x2=??x[?f?x], ?2f?x?y= ??x[?f?y], ?2f?y?x=??y[?f?x],?2f?y2=??y[?f?y].An alternative notation for each is

fxx,fyx,fxy,fxx,fyx,fxy, and fyy,fyy, respectively. Higher-order partial derivatives calculated with respect to different variables, such as fxyfxy and fyx,fyx, are commonly called mixed partial derivatives. Calculate all four second partial derivatives for the function f(x,y)=xe?3y+sin(2x?5y).f(x,y)=xe?3y+sin(2x?5y). To calculate ?2f/dx2?2f/dx2 and ?2f/

?y?x,?2f/?y?x, we first calculate ?f/?x:?f/?x: ? f ? x = e ?3 y + 2 cos ( 2 x ? 5 y ) . ? f ? x = e ?3 y + 2 cos ( 2 x ? 5 y ) . To calculate ?2f/dx2,?2f/dx2, differentiate ?f/?x?f/?x with respect to x:x: ? 2 f ? x 2 = ? ? x [ ? f ? x ] = ? ? x [ e ?3 y + 2 cos ( 2 x ? 5 y ) ] = ?4 sin ( 2 x ? 5 y ) . ? 2 f ? x 2 = ? ? x [ ? f ? x ] = ? ? x [ e ?3 y + 2 cos ( 2 x ? 5 y ) ] = ?4 sin ( 2

x ? 5 y ) . To calculate ?2f/?y?x,?2f/?y?x, differentiate ?f/?x?f/?x with respect to y:y: ? 2 f ? y ? x = ? ? y [ ? f ? x ] = ? ? y [ e ?3 y + 2 cos ( 2 x ? 5 y ) ] = ?3 e ?3 y + 10 sin ( 2 x ? 5 y ) . ? 2 f ? y ? x = ? ? y [ ? f ? x ] = ? ? y [ e ?3 y + 2 cos ( 2 x ? 5 y ) ] = ?3 e ?3 y + 10 sin ( 2 x ? 5 y ) . To calculate ?2f/?x?y?2f/?x?y and ?2f/dy2,?2f/dy2, first calculate ?f/

?y:?f/?y: ? f ? y = ?3 x e ?3 y ? 5 cos ( 2 x ? 5 y ) . ? f ? y = ?3 x e ?3 y ? 5 cos ( 2 x ? 5 y ) . To calculate ?2f/?x?y,?2f/?x?y, differentiate ?f/?y?f/?y with respect to x:x: ? 2 f ? x ? y = ? ? x [ ? f ? y ] = ? ? x [ ?3 x e ?3 y ? 5 cos ( 2 x ? 5 y ) ] = ?3 e ?3 y + 10 sin ( 2 x ? 5 y ) . ? 2 f ? x ? y = ? ? x [ ? f ? y ] = ? ? x [ ?3 x e ?3 y ? 5 cos ( 2 x ? 5 y ) ] = ?3 e

?3 y + 10 sin ( 2 x ? 5 y ) . To calculate ?2f/?y2,?2f/?y2, differentiate ?f/?y?f/?y with respect to y:y: ? 2 f ? y 2 = ? ? y [ ? f ? y ] = ? ? y [ ?3 x e ?3 y ? 5 cos ( 2 x ? 5 y ) ] = 9 x e ?3 y ? 25 sin ( 2 x ? 5 y ) . ? 2 f ? y 2 = ? ? y [ ? f ? y ] = ? ? y [ ?3 x e ?3 y ? 5 cos ( 2 x ? 5 y ) ] = 9 x e ?3 y ? 25 sin ( 2 x ? 5 y ) . Calculate all four second partial derivatives

for the function f(x,y)=sin(3x?2y)+cos(x+4y).f(x,y)=sin(3x?2y)+cos(x+4y). At this point we should notice that, in both Example 4.19 and the checkpoint, it was true that ?2f/?x?y=?2f/?y?x.?2f/?x?y=?2f/?y?x. Under certain conditions, this is always true. In fact, it is a direct consequence of the following theorem. Suppose that f(x,y)f(x,y) is defined on an

open disk DD that contains the point (a,b).(a,b). If the functions fxyfxy and fyxfyx are continuous on D,D, then fxy=fyx.fxy=fyx. Clairaut＊s theorem guarantees that as long as mixed second-order derivatives are continuous, the order in which we choose to differentiate the functions (i.e., which variable goes first, then second, and so on) does not matter.

It can be extended to higher-order derivatives as well. The proof of Clairaut＊s theorem can be found in most advanced calculus books. Two other second-order partial derivatives can be calculated for any function f(x,y).f(x,y). The partial derivative fxxfxx is equal to the partial derivative of fxfx with respect to x,x, and fyyfyy is equal to the partial

derivative of fyfy with respect to y.y. In Introduction to Differential Equations, we studied differential equations in which the unknown function had one independent variable. A partial differential equation is an equation that involves an unknown function of more than one independent variable and one or more of its partial derivatives. Examples of

partial differential equations are ut=c2(uxx+uyy)ut=c2(uxx+uyy) (4.17) (heat equation in two dimensions) utt=c2(uxx+uyy)utt=c2(uxx+uyy) (4.18) (wave equation in two dimensions) uxx+uyy=0uxx+uyy=0 (4.19) (Laplace＊s equation in two dimensions) In the first two equations, the unknown function uu has three independent variables〞

t,x,andyt,x,andy〞and cc is an arbitrary constant. The independent variables xandyxandy are considered to be spatial variables, and the variable tt represents time. In Laplace＊s equation, the unknown function uu has two independent variables xandy.xandy. Verify that u(x,y,t)=5sin(3羽x)sin(4羽y)cos(10羽t)u(x,y,t)=5sin(3羽x)sin(4羽y)cos(10羽t) is a solution

to the wave equation utt=4(uxx+uyy).utt=4(uxx+uyy). First, we calculate utt,uxx,utt,uxx, and uyy:uyy: u t t = ? ? t [ ? u ? t ] = ? ? t [ 5 sin ( 3 羽 x ) sin ( 4 羽 y ) ( ?10 羽 sin ( 10 羽 t ) ) ] = ? ? t [ ?50 羽 sin ( 3 羽 x ) sin ( 4 羽 y ) sin ( 10 羽 t ) ] = ?500 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) u x x = ? ? x [ ? u ? x ] = ? ? x [ 15 羽 cos ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10

羽 t ) ] = ?45 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) u y y = ? ? y [ ? u ? y ] = ? ? y [ 5 sin ( 3 羽 x ) ( 4 羽 cos ( 4 羽 y ) ) cos ( 10 羽 t ) ] = ? ? y [ 20 羽 sin ( 3 羽 x ) cos ( 4 羽 y ) cos ( 10 羽 t ) ] = ?80 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) . u t t = ? ? t [ ? u ? t ] = ? ? t [ 5 sin ( 3 羽 x ) sin ( 4 羽 y ) ( ?10 羽 sin ( 10 羽 t ) ) ] = ? ? t [ ?50 羽 sin ( 3 羽 x ) sin ( 4 羽 y )

sin ( 10 羽 t ) ] = ?500 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) u x x = ? ? x [ ? u ? x ] = ? ? x [ 15 羽 cos ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) ] = ?45 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) u y y = ? ? y [ ? u ? y ] = ? ? y [ 5 sin ( 3 羽 x ) ( 4 羽 cos ( 4 羽 y ) ) cos ( 10 羽 t ) ] = ? ? y [ 20 羽 sin ( 3 羽 x ) cos ( 4 羽 y ) cos ( 10 羽 t ) ] = ?80 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y )

cos ( 10 羽 t ) . Next, we substitute each of these into the right-hand side of Equation 4.20 and simplify: 4 ( u x x + u y y ) = 4 ( ?45 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) + ? 80 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) ) = 4 ( ?125 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) ) = ?500 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) = u t t . 4 ( u x x + u y y ) =

4 ( ?45 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) + ? 80 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) ) = 4 ( ?125 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) ) = ?500 羽 2 sin ( 3 羽 x ) sin ( 4 羽 y ) cos ( 10 羽 t ) = u t t . This verifies the solution. Verify that u(x,y,t)=2sin(x3)sin(y4)e?25t/16u(x,y,t)=2sin(x3)sin(y4)e?25t/16 is a solution to the heat equation

ut=9(uxx+uyy).ut=9(uxx+uyy). Since the solution to the two-dimensional heat equation is a function of three variables, it is not easy to create a visual representation of the solution. We can graph the solution for fixed values of t, which amounts to snapshots of the heat distributions at fixed times. These snapshots show how the heat is distributed over

a two-dimensional surface as time progresses. The graph of the preceding solution at time t=0t=0 appears in the following figure. As time progresses, the extremes level out, approaching zero as t approaches infinity. Figure 4.23 If we consider the heat equation in one dimension, then it is possible to graph the solution over time. The heat equation in

one dimension becomes ut=c2uxx,ut=c2uxx, where c2c2 represents the thermal diffusivity of the material in question. A solution of this differential equation can be written in the form um(x,t)=e?羽2m2c2tsin(m羽x)um(x,t)=e?羽2m2c2tsin(m羽x) (4.22) where mm is any positive integer. A graph of this solution using m=1m=1 appears in Figure 4.24,

where the initial temperature distribution over a wire of length 11 is given by u(x,0)=sin羽x.u(x,0)=sin羽x. Notice that as time progresses, the wire cools off. This is seen because, from left to right, the highest temperature (which occurs in the middle of the wire) decreases and changes color from red to blue. Figure 4.24 Graph of a solution of the heat

equation in one dimension over time. Figure 4.25 (a) William Thomson (Lord Kelvin), 1824-1907, was a British physicist and electrical engineer; (b) Kelvin used the heat diffusion equation to estimate the age of Earth (credit: modification of work by NASA). During the late 1800s, the scientists of the new field of geology were coming to the conclusion

that Earth must be ※millions and millions§ of years old. At about the same time, Charles Darwin had published his treatise on evolution. Darwin＊s view was that evolution needed many millions of years to take place, and he made a bold claim that the Weald chalk fields, where important fossils were found, were the result of 300300 million years of

erosion. At that time, eminent physicist William Thomson (Lord Kelvin) used an important partial differential equation, known as the heat diffusion equation, to estimate the age of Earth by determining how long it would take Earth to cool from molten rock to what we had at that time. His conclusion was a range of 20to40020to400 million years, but

most likely about 5050 million years. For many decades, the proclamations of this irrefutable icon of science did not sit well with geologists or with Darwin. Kelvin made reasonable assumptions based on what was known in his time, but he also made several assumptions that turned out to be wrong. One incorrect assumption was that Earth is solid

and that the cooling was therefore via conduction only, hence justifying the use of the diffusion equation. But the most serious error was a forgivable one〞omission of the fact that Earth contains radioactive elements that continually supply heat beneath Earth＊s mantle. The discovery of radioactivity came near the end of Kelvin＊s life and he

acknowledged that his calculation would have to be modified. Kelvin used the simple one-dimensional model applied only to Earth＊s outer shell, and derived the age from graphs and the roughly known temperature gradient near Earth＊s surface. Let＊s take a look at a more appropriate version of the diffusion equation in radial coordinates, which has

the form ?T?t=K[?2T?2r+2r?T?r].?T?t=K[?2T?2r+2r?T?r]. Here, T(r,t)T(r,t) is temperature as a function of rr (measured from the center of Earth) and time t.t. KK is the heat conductivity〞for molten rock, in this case. The standard method of solving such a partial differential equation is by separation of variables, where we express the solution as the

product of functions containing each variable separately. In this case, we would write the temperature as T(r,t)=R(r)f(t).T(r,t)=R(r)f(t). Substitute this form into Equation 4.13 and, noting that f(t)f(t) is constant with respect to distance (r)(r) and R(r)R(r) is constant with respect to time (t),(t), show that

1f?f?t=KR[?2R?r2+2r?R?r].1f?f?t=KR[?2R?r2+2r?R?r]. This equation represents the separation of variables we want. The left-hand side is only a function of tt and the right-hand side is only a function of r,r, and they must be equal for all values of randt.randt. Therefore, they both must be equal to a constant. Let＊s call that constant ?竹2.?竹2. (The

convenience of this choice is seen on substitution.) So, we have 1f?f?t=?竹2andKR[?2R?r2+2r?R?r]=?竹2.1f?f?t=?竹2andKR[?2R?r2+2r?R?r]=?竹2. Now, we can verify through direct substitution for each equation that the solutions are f(t)=Ae?竹2tf(t)=Ae?竹2t and R(r)=B(sin汐rr)+C(cos汐rr),R(r)=B(sin汐rr)+C(cos汐rr), where 汐=竹/K.汐=竹/K. Note that

f(t)=Ae+竹n2tf(t)=Ae+竹n2t is also a valid solution, so we could have chosen +竹2+竹2 for our constant. Can you see why it would not be valid for this case as time increases? Let＊s now apply boundary conditions. The temperature must be finite at the center of Earth, r=0.r=0. Which of the two constants, BB or C,C, must therefore be zero to keep RR

finite at r=0?r=0? (Recall that sin(汐r)/r↙汐=sin(汐r)/r↙汐= as r↙0,r↙0, but cos(汐r)/rcos(汐r)/r behaves very differently.) Kelvin argued that when magma reaches Earth＊s surface, it cools very rapidly. A person can often touch the surface within weeks of the flow. Therefore, the surface reached a moderate temperature very early and remained nearly

constant at a surface temperature Ts.Ts. For simplicity, let＊s set T=0atr=RET=0atr=RE and find 汐汐 such that this is the temperature there for all time t.t. (Kelvin took the value to be 300K＞80～F.300K＞80～F. We can add this 300K300K constant to our solution later.) For this to be true, the sine argument must be zero at r=RE.r=RE. Note that 汐汐 has

an infinite series of values that satisfies this condition. Each value of 汐汐 represents a valid solution (each with its own value for A).A). The total or general solution is the sum of all these solutions. At t=0,t=0, we assume that all of Earth was at an initial hot temperature T0T0 (Kelvin took this to be about 7000K.)7000K.) The application of this boundary

condition involves the more advanced application of Fourier coefficients. As noted in part b. each value of 汐n汐n represents a valid solution, and the general solution is a sum of all these solutions. This results in a series solution: T(r,t)=(T0RE羽)﹉n(?1)n?1ne?竹n2tsin(汐nr)r,where汐n=n羽/RE.T(r,t)=(T0RE羽)﹉n(?1)n?1ne?竹n2tsin(汐nr)r,where汐n=n羽/RE.

Note how the values of 汐n汐n come from the boundary condition applied in part b. The term ?1n?1n?1n?1n is the constant AnAn for each term in the series, determined from applying the Fourier method. Letting 汕=羽RE,汕=羽RE, examine the first few terms of this solution shown here and note how 竹2竹2 in the exponential causes the higher terms to

decrease quickly as time progresses: T(r,t)=T0RE羽r(e?K汕2t(sin汕r)?12e?4K汕2t(sin2汕r)+13e?9K汕2t(sin3汕r)?14e?16K汕2t(sin4汕r)+15e?25K汕2t(sin5汕r)...).T(r,t)=T0RE羽r(e?K汕2t(sin汕r)?12e?4K汕2t(sin2汕r)+13e?9K汕2t(sin3汕r)?14e?16K汕2t(sin4汕r)+15e?25K汕2t(sin5汕r)...). Near time t=0,t=0, many terms of the solution are needed for accuracy.

Inserting values for the conductivity KK and 汕=羽/RE汕=羽/RE for time approaching merely thousands of years, only the first few terms make a significant contribution. Kelvin only needed to look at the solution near Earth＊s surface (Figure 4.26) and, after a long time, determine what time best yielded the estimated temperature gradient known during

his era (1～F(1～F increase per 50ft).50ft). He simply chose a range of times with a gradient close to this value. In Figure 4.26, the solutions are plotted and scaled, with the 300?K300?K surface temperature added. Note that the center of Earth would be relatively cool. At the time, it was thought Earth must be solid. Figure 4.26 Temperature versus

radial distance from the center of Earth. (a) Kelvin＊s results, plotted to scale. (b) A close-up of the results at a depth of 4.0 mi 4.0 mi below Earth＊s surface. Epilog On May 20,1904,20,1904, physicist Ernest Rutherford spoke at the Royal Institution to announce a revised calculation that included the contribution of radioactivity as a source of Earth＊s

heat. In Rutherford＊s own words: ※I came into the room, which was half-dark, and presently spotted Lord Kelvin in the audience, and realised that I was in for trouble at the last part of my speech dealing with the age of the Earth, where my views conflicted with his. To my relief, Kelvin fell fast asleep, but as I came to the important point, I saw the old

bird sit up, open an eye and cock a baleful glance at me. Then a sudden inspiration came, and I said Lord Kelvin had limited the age of the Earth, provided no new source [of heat] was discovered. That prophetic utterance referred to what we are now considering tonight, radium! Behold! The old boy beamed upon me.§ Rutherford calculated an age for

Earth of about 500500 million years. Today＊s accepted value of Earth＊s age is about 4.64.6 billion years. Section 4.3 Exercises For the following exercises, calculate the partial derivative using the limit definitions only. 112. ?z?x?z?x for z=x2?3xy+y2z=x2?3xy+y2 113. ?z?y?z?y for z=x2?3xy+y2z=x2?3xy+y2 For the following exercises, calculate the

sign of the partial derivative using the graph of the surface. 114. f x ( 1 , 1 ) f x ( 1 , 1 ) 115. f x ( ?1 , 1 ) f x ( ?1 , 1 ) 116. f y ( 1 , 1 ) f y ( 1 , 1 ) 117. f x ( 0 , 0 ) f x ( 0 , 0 ) For the following exercises, calculate the partial derivatives. 118. ?z?x?z?x for z=sin(3x)cos(3y)z=sin(3x)cos(3y) 119. ?z?y?z?y for z=sin(3x)cos(3y)z=sin(3x)cos(3y) 120. ?z?x?z?x and

?z?y?z?y for z=x8e3yz=x8e3y 121. ?z?x?z?x and ?z?y?z?y for z=ln(x6+y4)z=ln(x6+y4) 122. Find fy(x,y)fy(x,y) for f(x,y)=exycos(x)sin(y).f(x,y)=exycos(x)sin(y). 123. Let z=exy.z=exy. Find ?z?x?z?x and ?z?y.?z?y. 124. Let z=ln(xy).z=ln(xy). Find ?z?x?z?x and ?z?y.?z?y. 125. Let z=tan(2x?y).z=tan(2x?y). Find ?z?x?z?x and ?z?y.?z?y. 126. Let

z=sinh(2x+3y).z=sinh(2x+3y). Find ?z?x?z?x and ?z?y.?z?y. 127. Let f(x,y)=arctan(yx).f(x,y)=arctan(yx). Evaluate fx(2,?2)fx(2,?2) and fy(2,?2).fy(2,?2). 128. Let f(x,y)=xyx?y.f(x,y)=xyx?y. Find fx(2,?2)fx(2,?2) and fy(2,?2).fy(2,?2). Evaluate the partial derivatives at point P(0,1).P(0,1). 129. Find ?z?x?z?x at (0,1)(0,1) for z=e?xcos(y).z=e?xcos(y).

130. Given f(x,y,z)=x3yz2,f(x,y,z)=x3yz2, find ?2f?x?y?2f?x?y and fz(1,1,1).fz(1,1,1). 131. Given f(x,y,z)=2sin(x+y),f(x,y,z)=2sin(x+y), find fx(0,羽2,?4),fx(0,羽2,?4), fy(0,羽2,?4),fy(0,羽2,?4), and fz(0,羽2,?4).fz(0,羽2,?4). 132. The area of a parallelogram with adjacent side lengths that are aandb,aandb, and in which the angle between these two sides is 牟,牟,

is given by the function A(a,b,牟)=basin(牟).A(a,b,牟)=basin(牟). Find the rate of change of the area of the parallelogram with respect to the following: Side a Side b Angle牟Angle牟 133. Express the volume of a right circular cylinder as a function of two variables: its radius rr and its height h.h. Show that the rate of change of the volume of the cylinder with

respect to its radius is the product of its circumference multiplied by its height. Show that the rate of change of the volume of the cylinder with respect to its height is equal to the area of the circular base. 134. Calculate ?w?z?w?z for w=zsin(xy2+2z).w=zsin(xy2+2z). Find the indicated higher-order partial derivatives. 135. fxyfxy for

z=ln(x?y)z=ln(x?y) 136. fyxfyx for z=ln(x?y)z=ln(x?y) 137. Let z=x2+3xy+2y2.z=x2+3xy+2y2. Find ?2z?x2?2z?x2 and ?2z?y2.?2z?y2. 138. Given z=extany,z=extany, find ?2z?x?y?2z?x?y and ?2z?y?x.?2z?y?x. 139. Given f(x,y,z)=xyz,f(x,y,z)=xyz, find fxyy,fyxy,fxyy,fyxy, and fyyx.fyyx. 140. Given f(x,y,z)=e?2xsin(z2y),f(x,y,z)=e?2xsin(z2y), show that

fxyy=fyxy.fxyy=fyxy. 141. Show that z=12(ey?e?y)sinxz=12(ey?e?y)sinx is a solution of the differential equation ?2z?x2+?2z?y2=0.?2z?x2+?2z?y2=0. 142. Find fxx(x,y)fxx(x,y) for f(x,y)=4x2y+y22x.f(x,y)=4x2y+y22x. 143. Let f(x,y,z)=x2y3z?3xy2z3+5x2z?y3z.f(x,y,z)=x2y3z?3xy2z3+5x2z?y3z. Find fxyz.fxyz. 144. Let

F(x,y,z)=x3yz2?2x2yz+3xz?2y3z.F(x,y,z)=x3yz2?2x2yz+3xz?2y3z. Find Fxyz.Fxyz. 145. Given f(x,y)=x2+x?3xy+y3?5,f(x,y)=x2+x?3xy+y3?5, find all points at which fx=fy=0fx=fy=0 simultaneously. 146. Given f(x,y)=2x2+2xy+y2+2x?3,f(x,y)=2x2+2xy+y2+2x?3, find all points at which ?f?x=0?f?x=0 and ?f?y=0?f?y=0 simultaneously. 147. Given

f(x,y)=y3?3yx2?3y2?3x2+1,f(x,y)=y3?3yx2?3y2?3x2+1, find all points on ff at which fx=fy=0fx=fy=0 simultaneously. 148. Given f(x,y)=15x3?3xy+15y3,f(x,y)=15x3?3xy+15y3, find all points at which fx(x,y)=fy(x,y)=0fx(x,y)=fy(x,y)=0 simultaneously. 149. Show that z=exsinyz=exsiny satisfies the equation ?2z?x2+?2z?y2=0.?2z?x2+?2z?y2=0.

150. Show that f(x,y)=ln(x2+y2)f(x,y)=ln(x2+y2) solves Laplace＊s equation ?2z?x2+?2z?y2=0.?2z?x2+?2z?y2=0. 151. Show that z=e?tcos(xc)z=e?tcos(xc) satisfies the heat equation ?z?t=c2?2z?x2?z?t=c2?2z?x2 152. Find lim忖x↙0f(x+忖x, y)?f(x,y)忖xlim忖x↙0f(x+忖x, y)?f(x,y)忖x for f(x,y)=?7x?2xy+7y.f(x,y)=?7x?2xy+7y. 153. Find lim忖y↙0f(x,y+忖y)

?f(x,y)忖ylim忖y↙0f(x,y+忖y)?f(x,y)忖y for f(x,y)=?7x?2xy+7y.f(x,y)=?7x?2xy+7y. 154. Find lim忖x↙0忖f忖x=lim忖x↙0f(x+忖x,y)?f(x,y)忖xlim忖x↙0忖f忖x=lim忖x↙0f(x+忖x,y)?f(x,y)忖x for f(x,y)=x2y2+xy+y.f(x,y)=x2y2+xy+y. 155. Find lim忖x↙0忖f忖x=lim忖x↙0f(x+忖x,y)?f(x,y)忖xlim忖x↙0忖f忖x=lim忖x↙0f(x+忖x,y)?f(x,y)忖x for f(x,y)=sin(xy).f(x,y)=sin(xy). 156.

The function P(T,V)=nRTVP(T,V)=nRTV gives the pressure at a point in a gas as a function of temperature TT and volume V.V. The letters nandRnandR are constants. Find ?P?V?P?V and ?P?T,?P?T, and explain what these quantities represent. 157. The equation for heat flow in the xy-planexy-plane is ?f?t=?2f?x2+?2f?y2.?f?t=?2f?x2+?2f?y2. Show that

f(x,y,t)=e?2tsinxsinyf(x,y,t)=e?2tsinxsiny is a solution. 158. The basic wave equation is ftt=fxx.ftt=fxx. Verify that f(x,t)=sin(x+t)f(x,t)=sin(x+t) and f(x,t)=sin(x?t)f(x,t)=sin(x?t) are solutions. 159. The law of cosines can be thought of as a function of three variables. Let x,y,x,y, and 牟牟 be two sides of any triangle where the angle 牟牟 is the included

angle between the two sides. Then, F(x,y,牟)=x2+y2?2xycos牟F(x,y,牟)=x2+y2?2xycos牟 gives the square of the third side of the triangle. Find ?F?牟?F?牟 and ?F?x?F?x when x=2,y=3,x=2,y=3, and 牟=羽6.牟=羽6. 160. Suppose the sides of a rectangle are changing with respect to time. The first side is changing at a rate of 22 in./sec whereas the second side

is changing at the rate of 44 in/sec. How fast is the diagonal of the rectangle changing when the first side measures 1616 in. and the second side measures 2020 in.? (Round answer to three decimal places.) 161. A Cobb-Douglas production function is f(x,y)=200x0.7y0.3,f(x,y)=200x0.7y0.3, where xandyxandy represent the amount of labor and capital

available. Let x=500x=500 and y=1000.y=1000. Find 汛f汛x汛f汛x and 汛f汛y汛f汛y at these values, which represent the marginal productivity of labor and capital, respectively. 162. The apparent temperature index is a measure of how the temperature feels, and it is based on two variables: h,h, which is relative humidity, and t,t, which is the air temperature.

A=0.885t?22.4h+1.20th?0.544.A=0.885t?22.4h+1.20th?0.544. Find ?A?t?A?t and ?A?h?A?h when t=20～Ft=20～F and h=0.90.h=0.90.

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