Implicit differentiation practice problems with answers pdf

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Implicit differentiation practice problems with answers pdf

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation. Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best

views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width. For problems 1 ? 3 do each of the following. Find \(y'\) by solving the equation for y and differentiating directly. Find \(y'\) by

implicit differentiation. Check that the derivatives in (a) and (b) are the same. For problems 4 ? 9 find \(y'\) by implicit differentiation. \(2{y^3} + 4{x^2} - y = {x^6}\) Solution \(7{y^2} + \sin \left( {3x} \right) = 12 - {y^4}\) Solution \({{\bf{e}}^x} - \sin \left( y \right) = x\) Solution \(4{x^2}{y^7} - 2x = {x^5} + 4{y^3}\) Solution \(\cos \left( {{x^2} + 2y} \right) + x\,

{{\bf{e}}^{{y^{\,2}}}} = 1\) Solution \(\tan \left( {{x^2}{y^4}} \right) = 3x + {y^2}\) Solution For problems 10 & 11 find the equation of the tangent line at the given point. \({x^4} + {y^2} = 3\) at \(\left( {1,\, - \sqrt 2 } \right)\). Solution \({y^2}{{\bf{e}}^{2x}} = 3y + {x^2}\) at \(\left( {0,3} \right)\). Solution For problems 12 & 13 assume that \(x = x\left( t \right)\), \(y = y\left( t

\right)\) and \(z = z\left( t \right)\) and differentiate the given equation with respect to t. \({x^2} - {y^3} + {z^4} = 1\) Solution \({x^2}\cos \left( y \right) = \sin \left( {{y^3} + 4z} \right)\) Solution If a function is described by the equation \(y = f\left( x \right)\) where the variable \(y\) is on the left side, and the right side depends only on the independent variable \(x\), then

the function is said to be given explicitly. For example, the following functions are defined explicitly: \[ {y = \sin x,\;\;\;}\kern-0.3pt {y = {x^2} + 2x + 5,\;\;\;}\kern-0.0pt {y = \ln \cos x.} \] In many problems, however, the function can be defined in implicit form, that is by the equation \[F\left( {x,y} \right) = 0.\] Of course, any explicit function can be written in an implicit

form. So the above functions can be represented as \[ {y ? \sin x = 0,\;\;\;}\kern-0.0pt {y ? {x^2} ? 2x ? 5 = 0,\;\;\;}\kern-0.0pt {y ? \ln \cos x = 0.} \] The inverse transformation cannot be always performed. There are often functions defined by an implicit equation that cannot be resolved with respect to the variable \(y.\) Examples of such implicit functions are \[

{{x^3} + {y^3} ? 3{x^2}{y^5} = 0,\;\;\;}\kern-0.3pt {\frac{{x ? y}}{{\sqrt {{x^2} + {y^2}} }} ? 4x{y^2} = 0,\;\;\;}\kern-0.3pt {xy ? \sin \left( {x + y} \right) = 0.} \] The good news is that we do not need to convert an implicitly defined function into an explicit form to find the derivative \(y'\left( x \right).\) If \(y\) is defined implicitly as a function of \(x\) by an equation \(F\left(

{x,y} \right) = 0,\) we proceed as follows: Differentiate both sides of the equation with respect to \(x\), assuming that \(y\) is a differentiable function of \(x\) and using the chain rule. The derivative of zero (in the right side) will also be equal to zero. Note: If the right side is different from zero, that is the implicit equation has the form \[f\left( {x,y} \right) = g\left( {x,y}

\right),\] then we differentiate the left and right side of the equation.Solve the resulting equation for the derivative \(y'\left( x \right)\). In the examples below find the derivative of the implicit function. Solved Problems Click or tap a problem to see the solution. Find the derivative of the function given by the equation \({y^2} = 2px,\) where \(p\) is a parameter.

Differentiate implicitly the function \(y\left( x \right)\) given by the equation \(y = \cos \left( {x + y} \right).\) Calculate the derivative at the point \(\left( {0,0} \right)\) of the function given by the equation \(x = y ? 2\sin y.\) Find the equation of the tangent line to the curve \({x^4} + {y^4} = 2\) at the point \(\left( {1,1} \right).\) Calculate the derivative of the function \

(y\left( x \right)\) given by the equation \({x^2} + 2xy + 2{y^2} = 1\) under condition \(y = 1.\) Given the equation of a circle \({x^2} + {y^2} = {r^2}\) of radius \(r\) centered at the origin. Find the derivative \(y'\left( x \right).\) \[{x^2} + {y^2} ? 2x ? 4y = 4\] \[{x^3} + 2{y^3} + y{x^2} = 3\] Calculate the derivative at the point \(\left( {0,0} \right)\) of the function given by the

equation \[{x^5} + {y^5} ? 2x + 2y = 0.\] \[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\] \[{3^x} + {3^y} = {3^{x + y}}\] Find the derivative of the astroid \({x^{\large\frac{2}{3}ormalsize}} + {y^{\large\frac{2}{3}ormalsize}} = {a^{\large\frac{2}{3}ormalsize}}.\) Find the derivative \(y'\left( x \right)\) of the function that describes the general equation of a second order

curve: \[{A{x^2} + 2Bxy + C{y^2} + 2Dx }+{ 2Ey + F = 0.}\] Find the value of the derivative \(y'\left( x \right)\) at \(x = 0\) for the function given by \({e^y} + xy = e.\) \[x\sin y + y\cos x = 0\] \[y = \sin \left( {x ? y} \right)\] \[y = \sin \left( {x + y} \right)\] \[{x^2} ? \sin \left( {xy} \right) = 0\] \[{{2^{\large\frac{x}{y}ormalsize}} = \frac{{{x^2}}}{{{y^2}}}\;\;}\kern-0.3pt{\left( {y e 0}

\right).}\] \[{{x^2} + y + \ln \left( {x + y} \right) = 0,\;\;\;}\kern-0.3pt{\left( {x + y \gt 0} \right).}\] Find the derivative at the point \(x = 2\) for the function given by the equation \[{{x^2} + {y^2} + 2x ? xy }+{ 5y ? 2 = 0.}\] \[{\frac{y}{x} = \ln \left( {xy} \right)\;\;\;}\kern-0.3pt{\left( {xy \gt 0} \right).}\] Calculate the derivative at the point \(\left( {1,1} \right)\) of the function given

by the equation \(x ? y = \ln \left( {xy} \right)\) where \(xy \gt 0.\) Find the derivative of the function defined by the equation \(x + y = \arctan \left( {xy} \right).\) \[x ? y + \arctan y = 0\] Find the derivative of the function given by the equation \({y^2} = 2px,\) where \(p\) is a parameter. Solution. This equation is the canonical equation of a parabola. Differentiating the

left and right sides with respect to \(x\), we have: \[ {{\left( {{y^2}} \right)^\prime } = {\left( {2px} \right)^\prime },\;\;}\Rightarrow {2yy' = 2p,\;\;}\Rightarrow {y' = \frac{p}{y},\;\;\;}\kern-0.3pt {\text{where}\;\;y e 0.} \] Differentiate implicitly the function \(y\left( x \right)\) given by the equation \(y = \cos \left( {x + y} \right).\) Solution. Differentiate both sides with respect to \

(x:\) \[ {\frac{{dy}}{{dx}} = \frac{d}{{dx}}\cos \left( {x + y} \right),\;\;}\Rightarrow {y' = ? \sin \left( {x + y} \right) \cdot \left( {1 + y'} \right),}\Rightarrow {y' = ? \sin \left( {x + y} \right) }-{ y'\sin \left( {x + y} \right),}\Rightarrow {y'\left( {1 + \sin \left( {x + y} \right)} \right) }={ ? \sin \left( {x + y} \right),} \] which results in \[{y' }={ ? \frac{{\sin \left( {x + y} \right)}}{{1 + \sin \left( {x

+ y} \right)}}.}\] Calculate the derivative at the point \(\left( {0,0} \right)\) of the function given by the equation \(x = y ? 2\sin y.\) Solution. We differentiate both sides of the equation with respect to \(x\) and solve for \(y^\prime:\) \[{x^\prime = y^\prime ? \left( {2\sin y} \right)^\prime,}\;\; \Rightarrow {1 = y^\prime ? 2\cos y \cdot y^\prime,}\;\; \Rightarrow {y^\prime =

\frac{1}{{1 ? 2\cos y}}.}\] Substitute the coordinates \(\left( {0,0} \right):\) \[{y^\prime\left( {0,0} \right) = \frac{1}{{1 ? 2\cos 0}} }={ \frac{1}{{1 ? 2 \cdot 1}} }={ ? 1.}\] Find the equation of the tangent line to the curve \({x^4} + {y^4} = 2\) at the point \(\left( {1,1} \right).\) Solution. Differentiate both sides of the equation with respect to \(x:\) \[ {\frac{d}{{dx}}\left( {{x^4} +

{y^4}} \right) = \frac{d}{{dx}}\left( 2 \right),\;\;}\Rightarrow {4{x^3} + 4{y^3}y' = 0,\;\;}\Rightarrow {{x^3} + {y^3}y' = 0.} \] Then \(y' = ? {\large\frac{{{x^3}}}{{{y^3}}}ormalsize}\). At the point \(\left( {1,1} \right)\) we have \(y'\left( 1 \right) = ? 1.\) Hence, the equation of the tangent line is given by \[ {\frac{{x ? 1}}{{y ? 1}} = ? 1\;\;\;}\kern-0.3pt {\text{or}\;\;x + y = 2.} \]

Calculate the derivative of the function \(y\left( x \right)\) given by the equation \({x^2} + 2xy + 2{y^2} = 1\) under condition \(y = 1.\) Solution. We differentiate both sides of the equation implicitly with respect to \(x\) (we consider the left side as a composite function and use the chain rule): \[ {\frac{d}{{dx}}\left( {{x^2} + 2xy + 2{y^2}} \right) = \frac{d}{{dx}}\left( 1

\right),}\Rightarrow {2x + 2\left( {y + xy'} \right) + 4yy' = 0,}\Rightarrow {x + y + xy' + 2yy' = 0.} \] When \(y = 1,\) the original equation becomes \[ {{x^2} + 2x + 2 = 1,\;\;}\Rightarrow {{x^2} + 2x + 1 = 0,\;\;}\Rightarrow {{\left( {x + 1} \right)^2} = 0,\;\;}\Rightarrow {x = ? 1.} \] Substituting the values \(x = -1\) and \(y = 1\), we obtain: \[{ ? 1 + 1 ? y' + 2y' }={ 0.}\] It

follows from here that \(y' = 0\) at \(y = 1.\) Given the equation of a circle \({x^2} + {y^2} = {r^2}\) of radius \(r\) centered at the origin. Find the derivative \(y'\left( x \right).\) Solution. Differentiate both sides of the equation with respect to \(x:\) \[{\frac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \frac{d}{{dx}}\left( {{r^2}} \right),\;\;}\Rightarrow{2x + 2yy' = 0,\;\;}\Rightarrow{x

+ yy' = 0,\;\;}\Rightarrow{yy' = ? x,\;\;}\Rightarrow{y' = ? \frac{x}{y}.}\] (1) Find the derivative of y = x cos x Solution(2) Find the derivative of y = x log x + (log x)x Solution(3) Find the derivative of (xy) = e x - y Solution(4) Find the derivatives of the followingxy = yx Solution(5) Find the derivatives of the followingy = (cos x) log x

Solution(6) Find the derivatives of the following(x2/a2) + (y2/b2) = 1 Solution(7) Find the derivatives of the following(x2 + y2) = tan-1(y/x) Solution(8) Differentiate the followingtan (x + y) + tan (x - y) = x

Solution(9) Differentiate the followingIf cos (xy) = x, show that dy/dx = -(1+ysin(xy))/x sin(xy)

Solution(10) Differentiate the

followingtan-1[(1 - cos x)/(1+cosx)]

Solution(11) Differentiate the followingtan-1[6x/1-9x2]

Solution(12) Differentiate the followingcos (2tan-1[(1-x)/(1+x)])

Solution(13) Differentiate the followingx = a cos3t, y = a sin3t

Solution(14) Differentiate the followingx = a (cos t + t sin t) ; y = a (sin t - t cos t)

Solution(15) Differentiate the followingx = (1-t2)/(1+t2) ; y = 2t/(1+t2)

Solution(16) Differentiate the followingcos-1(1 -x2)/(1+x2)

Solution(17) Differentiate the followingsin-1(3x - 4x3)

Solution(18) Differentiate the followingtan-1[(cos x + sin x) / (cos x - sin x)]

Solution(19) Find the derivative of sin x2 with respect to x2 Solution(20) Find

the derivative of sin-1(2x / (1 + x2)) with respect to tan-1 x

Solution(21) Differentiate the followingIf u = tan-1 [(1+x2) - 1]/x and v = tan-1x, find du/dv Solution(22) Find the derivative with tan-1 (sin x/(1 + cos x)) with respect to tan-1 (cos x/(1 + sin x)) Solution(23) If y = sin-1x then find y''

Solution(24) If y = e^(tan-1x) show that (1+ x2 ) y''

+ (2x -1) y' = 0.

Solution(25) If y = sin-1 x/(1-x2) show that (1-x2)y2 - 3xy1 - y = 0

Solution(26) If x = a ( + sin ), y = a (1 - cos ) then prove that at = /2, y'' = 1/a

Solution(27) If sin y = x sin(a + y), then prove that dy/dx = sin2 (a + y)/sin a , a n

Solution(28) If y = (cos-1x)2 prove that (1 -x2)(d2y/dx2) - x (dy/dx) - 2 = 0

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