Differentiation of Exponential Functions

Formulas and examples of the derivatives of exponential functions, in calculus, are presented. Several examples, with detailed solutions, involving products, sums and quotients of exponential functions are examined.

First Derivative of Exponential Functions to any Base

The derivative of f(x) = b^x is given by
f '(x) = b^x ln b
Note: if f(x) = e^x , then f '(x) = e^x

Example 1

Find the derivative of f(x) = 2^x

Solution to Example 1

Apply the formula above to obtain
f '(x) = 2^x ln 2

Example 2

Find the derivative of f(x) = 3^x + 3x²

Solution to Example 2

Let g(x) = 3^x and h(x) = 3x², function f is the sum of functions g and h: f(x) = g(x) + h(x). Use the sum rule, f '(x) = g '(x) + h '(x), to find the derivative of function f
f '(x) = 3^x ln 3 + 6x

Example 3

Find the derivative of f(x) = e^x / ( 1 + x )

Solution to Example 3

Let g(x) = e^x and h(x) = 1 + x, function f is the quotient of functions g and h: f(x) = g(x) / h(x). Hence we use the quotient rule, f '(x) = [ h(x) g '(x) - g(x) h '(x) ] / h(x)², to find the derivative of function f.
g '(x) = e^x
h '(x) = 1
f '(x) = [ h(x) g '(x) - g(x) h '(x) ] / h(x)²
= [ (1 + x)(e^x) - (e^x)(1) ] / (1 + x)²
Multiply factors in the numerator and simplify
f '(x) = x e^x / (1 + x)²

Example 4

Find the derivative of f(x) = e^{2x + 1}

Solution to Example 4

Let u = 2x + 1 and y = e^u, Use the chain rule to find the derivative of function f as follows.
f '(x) = (dy / du) (du / dx)
dy / du = e^u and du / dx = 2
f '(x) = (e^u)(2) = 2 e^u
Substitute u = 2x + 1 in f '(x) above
f '(x) = 2 e^{2x + 1}

Exercises

Find the derivative of each function.
1 - f(x) = e^x 2^x
2 - g(x) = 3^x - 3x³
3 - h(x) = e^x / (2x - 3)
4 - j(x) = e^{(x² + 2)}

Solutions to the Above Exercises

1 - f '(x) = e^x 2^x ( ln 2 + 1)
2 - g '(x) = 3^x ln 3 - 9x²
3 - h '(x) = e^x(2x -5) / (2x - 3)²
4 - j '(x) = 2x e^{(x² + 2)}

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