Common Derivatives and Integrals

[Pages:12]Common Derivatives and Integrals

Derivative Rules:

1. Constant Multiple Rule d [cu] = cu , where c is a constant.

dx

2. Sum and Difference Rule d [u ? v] = u ? v

dx

3. Product Rule d [uv] = uv + vu

dx

4. Quotient Rule

d u dx v

=

vu - uv v2

5. Constant Rule, d [c] = 0

dx

[ ] 6. Power Rule d u n = nu n-1u dx

7. Power Rule d [x] = 1

dx

[ ] 8. Derivative Involving Absolute Value d u = u (u),u 0

dx

u

9. Derivative of the Natural Logarithmic Function d [ln u] = u

dx

u

[ ] 10. Derivative of Natural Exponential Function d eu = euu dx

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Common Derivatives and Integrals Reviewed June 2008

( )( ) Example 1: Find the derivative of f (x) = 4x - 3x2 3 + 2x2

Since there are two polynomials multiplied by each other, apply the third derivative rule, the Product Rule, to the problem.

This is the result of the Product Rule:

[ ] [ ] ( ) ( ) f (x) = 4x - 3x2 d 3 + 2x2 + 3 + 2x2 d 4x - 3x2

dx

dx

Now, take the derivative of each term inside of the brackets. Multiple derivative rules are used, including the Sum and Difference Rule, Constant Rule, Constant Multiple Rule, and Power Rule. When applied, the result is:

f (x) = (4x - 3x2 )(0 + 4x) + (3 + 2x2 )(4 - 6x)

Simplify:

f (x) = (4x - 3x2 )(4x) + (3 + 2x2 )(4 - 6x)

Multiply the polynomials by each other:

( ) ( ) f (x) = 16x2 -12x3 + 12 + 8x2 -18x -12x3

f (x) = 16x2 -12x3 + 12 + 8x2 -18x -12x3

Combine like terms to get a simplified answer:

f (x) = -24x3 + 24x2 -18x + 12

Integral Formulas:

Indefinite integrals have +C as an arbitrary constant.

1. kf (u)du = k f (u)du , where k is a constant.

2. [ f (u) ? g(u)]du = f (u)du ? g(u)du

3. du = u + C

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Common Derivatives and Integrals

 4. u ndu = u n+1 + C, n -1

n +1

5.

du u

=

ln u

+C

6. eu du = eu + C

( ) Example 2: Evaluate 4x2 - 5x3 + 12 dx

To evaluate this problem, use the first four Integral Formulas. First, use formula 2 to make the large integral into three smaller integrals:

( ) 4x2 - 5x3 + 12 dx = 4x2dx - 5x3dx + 12dx

Second, pull out the constants by using formula 1:

= 4 x2dx - 5 x3dx + 12 dx

Now find each integral using formulas 3 and 4:

=

4

x 2

2 +1

+1

-

5

x 3

3+1

+1

+

12(

x)

+

C

Although three integrals have been removed, only one constant C is needed because C represents all unknown constants. Therefore multiple C's can be combined into just one C. To get the final answer, simplify the expression:

=

4

x3 3

-

5

x4 4

+

12x

+

C

= 4 x3 - 5 x 4 + 12x + C 34

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Common Derivatives and Integrals

Derivatives Rules for Trigonometric Functions:

1. d [sin(u)] = (cos(u))u

dx

2. d [cos(u)] = -(sin(u))u

dx

3. d [tan(u)] = (sec2 (u))u dx

4. d [cot(u)] = -(csc2 (u))u dx

5. d [sec(u)] = (sec(u)tan(u))u

dx

6. d [csc(u)] = -(csc(u)cot(u))u

dx

Example 3: Find the derivative of

f

(x)

=

sin(x) cos(x)

When finding the derivatives of trigonometric functions, non-trigonometric derivative rules are often incorporated, as well as trigonometric derivative rules. Looking at this function, one can see that the function is a quotient. Therefore, use derivative rule 4 on page 1, the Quotient Rule, to start this problem:

(cos(x)) d [sin(x)]- (sin(x)) d [cos(x)]

f (x) =

dx

dx

cos2 (x)

Now use trigonometric derivative rules 1 and 2 to get:

f

(x)

=

(cos(x))(cos(x)) - (sin(x))(- cos2 (x)

sin(x))

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Common Derivatives and Integrals

Once the multiplication has been completed in the numerator of the fraction, the result is:

f

(x)

=

(sin

2

(x) +

cos 2

cos 2

(x)

(x))

Remember that sin 2 (x) + cos2 (x) = 1; therefore, substitute 1 for sin2 (x) + cos2 (x) in the

answer. The final result is:

f

(x)

=

1 cos 2

(x)

f (x) = sec2 (x)

Integrals of Trigonometric Functions:

1. sin(u)du = - cos(u) + C

2. cos(u)du = sin(u) + C

3 tan(u)du = - ln cos(u) + C

4. cot(u)du = ln sin(u) + C

5. sec(u)du = ln sec(u) + tan(u) + C

6. csc(u)du = - ln csc(u) + cot(u) + C

7. sec2 (u)du = tan(u) + C

8. csc2 (u)du = - cot(u) + C

9. sec(u)tan(u)du = sec(u) + C

10. csc(u)cot(u)du = - csc(u) + C

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Common Derivatives and Integrals

Example 4: Evaluate (cos(x) - tan(x))dx

Normal integration formulas are often used in addition to trigonometric formulas when doing trigonometric integration. For example, in this problem use integration formula 2:

(cos(x) - tan(x))dx = cos(x)dx - tan(x)dx

With the two smaller integrals, use trigonometric integration formulas 2 and 3 to find the solution:

= sin(x) - (- ln cos(x))+ C

Simplify:

= sin(x) + ln cos(x) + C

Special Differentiation Rules:

Chain Rule: In certain situations, there may be a differentiable function of u, such as

y = f (u), and u = g(x) , where g(x) is a differentiable function of x. If this is the case, then y = f (g(x)) is a differentiable function of x. To take the derivative of the composite function y = f (g(x)), use the formula:

dy = dy du dx du dx

This formula can be rewritten as:

d [ f (g(x))] = f (g(x))g(x)

dx

( ) Example 5: Find dy

of the function y =

3+ x4

2

.

dx

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Common Derivatives and Integrals

( ) Use the formula dy = dy du to find this derivative. Now 3 + x4 = u, and the result is the dx du dx function y = u 2 .

Using the Chain Rule, we have

( )( ) ( ) dy = 2 3 + x4 4x3 = 8x3 3 + x4 , where

dx

( ) dy = 2 3 + x4 And du = 4x3

du

dx

Special Integration Formulas:

U-Substitution: Some integrals cannot be solved by using only the basic integration formulas. In some of these cases, one can use a process called u-substitution. This process helps simplify a problem before solving it.

Example 6:

Solve

(2

+

3x

)

1 2

dx

Referring to the given integral formulas, there are none that are able to solve this integral in its current form. When one comes to an integral in a form like this, it may be possible to simplify the integral to a form that is solvable by the given formulas. For this integral, usubstitution may be used. :

Using U- Substitution shows that:

u = 2 + 3x

du = (2 + 3x) = 3

dx

dx du = 3dx dx

du = 3dx

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Common Derivatives and Integrals

1 Having found that 2 + 3x = u, and that dx = du, substitute both into the original integral,

3

1

1 2

2 + 3xdx . The result is

u du , which can be solved by using integral formulas 1 and 4. 3

First, use formula 1, to find that:

1

1 2

1 1

= u du = u 2 du

3

3

Integration formula 4 shows that:

1 1

= u 2 du 3

=

1

3

1 +1

u2

+C

1 2

+

1

=

1

3

1+2

u2 2

+C

1 2

+

2 2

=

1

u

3 2

3

3 2

+C

=

1

2

u

3 2

+C

33

=

2 3 u2

+C

9

After solving the integral for u, substitute u with 2 + 3x to get the final answer:

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Common Derivatives and Integrals

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