DERIVATIVES OF EXPONENTIAL FUNCTIONS - Differentiation of Transcendental Functions - The Calculus Primer

The Calculus Primer (2011)

Part V. Differentiation of Transcendental Functions

Chapter 17. DERIVATIVES OF EXPONENTIAL FUNCTIONS

5—6. Differentiating the Simple Exponential Function. Let us consider first the simple exponential function y = av, where a is a constant greater than zero (that is, positive), and v is a function of x. Taking the logarithm of both sides to the base e, we have:

log y = v log a,

images

Differentiating with respect to y:

images

Applying the formula for the derivative of an inverse function, §5—3, equation [2]:

images

Now we apply the formula for the derivative of a function of a function, §5—1, equation [1], since v is a function of x; that is

images

Substituting the value of images from (1) above, we get:

images

As a special case, when a = e, log a = log e = 1; hence

images

RULE. The derivative of a constant with a variable exponent is equal to the product of the natural logarithm of the constant, the constant with the variable exponent, and the derivative of the exponent.

EXAMPLE 1.Find images (a4x+1).

Solution.images (a4x+1) = log a(a4x+1)(4),or4 log a(a4x+1).

EXAMPLE 2.Differentiate y = e−2x.

Solution.images = e−2x(−2) = −2e−2x.

EXAMPLE 3.Differentiate s = et3+1.

Solution.images

EXAMPLE 4.Find images in the equation ρ = aθ2.

Solution.images = log a·aθ2·2θ.

EXAMPLE 5.Differentiate y = elog x.

Solution.images

But elog x = x, by the definition of a logarithm.

Thereforeimages

EXAMPLE 6.Differentiate y = xe2/x.

Solution.

images

EXERCISE 5—2

Differentiate:

1. y = k4x

images

3. y = ekx

4. y = kex

5. y = e5−3t

images

images

11. ρ = aθ

12. ρ = alogθ

13. ρ = e

14. y = e1/x

15. y = xex

16. y = x2e2x

17. y = xn + nx

18. y = axxa

19. y = ex(x2 − 2x + 2)

20. y = ex(x − 1)

5—7. Differentiating the General Exponential Function. We now consider the more general exponential function y = uv, that is, a function raised to a variable power instead of a constant raised to a variable power. The only restriction is that u shall assume only positive values.

Lety = uv.(1)

Take the logarithm of both sides to the base e:

loge y = v loge u,

or, by the definition of a logarithm,

y = ev log u.(2)

Differentiating equation (2) by formula [4a]:

images

But ev log u = y from equation (2); and y = uv from equation (1); hence ev log u = uv.

Also, by differentiating as a product,

images

Making these substitutions in equation (3), we get:

images

RULE. The derivative of a function with a variable exponent is equal to the sum of the two results obtained by first regarding the exponent as a constant and differentiating, and then by regarding the function as a constant and differentiating.

EXAMPLE 1.Differentiate y = xx.

Solution.

images

EXAMPLE 2.Differentiate images

Solution.Here u = x, v = ex.

images

EXAMPLE 3.Prove that the formula images holds for all values of the constant n by setting v = n in formula [5].

Solution.If v = n, we have, from [5]:

images

EXERCISE 5—3

Differentiate:

1. y = xx+1

2. y = (x2 + 1)x

3. y = (x3)x

4. y = x2x

5. y = (2x)x3

6. y = xx2

7. y = x1/x

8. y = xlog x

5—8. Logarithmic Differentiation. When differentiating logarithmic and exponential functions, it is often more convenient to transform the given expression by making use of the properties of logarithms, namely:

log AB = log A + log B;log An = n log A;

images

This is known as logarithmic differentiation, and is now illustrated.

EXAMPLE 1.Differentiate images

Solution.First write

y = imageslog(x3 + 2),

thus eliminating the radical.

Thenimages

EXAMPLE 2.Differentiate images

Solution.First write

y = log x2 − log (x + 1).

images

EXAMPLE 3.Differentiate images

Solution.First write

y = images[log (x2a) − log (x2 + a)].

Thenimages

EXAMPLE 4.Differentiate images

Solution.First take the logarithm of both sides:

log y = log (x + 3) + log (x + 2) − log (x + 1).

Then, differentiating both sides with respect to x:

images

EXAMPLE 5.Differentiate y = xx+1.

Solution.First take the logarithm of both sides:

log y = (x+ 1) log x.

Differentiating both sides with respect to x:

images

EXAMPLE 6.Differentiate images

Solution.Taking the logarithm of both sides:

log y = x2 log x.

Differentiating both sides with respect to x:

images · images = x2 · images + log x·2x

= x(1 + 2 log x).

images = xx2+1(1 + 2 log x).

EXERCISE 5—4

Differentiate, using the method of logarithmic differentiation:

images

2. y = xxn

3. y = exx

images

images