DIFFERENTIATION OF EXPONENTIAL FUNCTIONS

 

DERIVATIVE OF THE EXPONENTIAL FUNCTION

The derivative of e^x is quite remarkable. The expression for the derivative is the same as the expression that we started with; that is, e^x!

$What does this mean?$ It means the slope is the same as the function value (the y-value) for all points on the graph.

Example: Let's take the example when x = 2. At this point, the y-value is e² ≈ 7.39.
Since the derivative of e^x is e^x, then the slope of the tangent line at x = 2 is also e² ≈ 7.39.
We can see that it is true on the graph:

Let's now see if it is true at some other values of x.

We can see that at x = 4, the y-value is 54.6 and the slope of the tangent (in red) is also 54.6.
At x = 5, the y-value is 148.4, as is the value of the derivative and the slope of the tangent (in green).

DIFFERENTIATION OF EXPONENTIAL FUNCTIONS

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 + 3x 2 Solution to Example 2: ·         Let g(x) = 3 x and h(x) = 3x 2, 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 - 3x 3 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 - 9x 2 3 - h '(x) = e x(2x -5) / (2x - 3) 2 4 - j '(x) = 2x e (x2 + 2)

   

Solve the following problem

Differentiate y=e^(ax+b) where a and b are constants

Let u = ax + b and y = e ^u

 Use the chain rule to find the derivative of function as follows 

 dy/dx = (dy / du) (du / dx) 

dy / du = e ^u and du / dx = a  ( Since the derivative of the constant b is 0)

Hence dy/dx = (e ^u)(a) = a e ^u

Substitute u = ax + b in dy/dx above 

dy/dx = a e ^ax + b