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Saturday, February 20, 2016

DIFFERENTIATION OF EXPONENTIAL FUNCTIONS

 

DERIVATIVE OF THE EXPONENTIAL FUNCTION

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


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