Derivative of ex is ex. The derivative of ex means the change in the exponential with respect to x. It is denoted by d(ex)/ dx. It is written as f'(x) = ex, where 'e' is the Euler's number and its value is approximately 2.718.
Formula for the derivative of ex is given below,
d(ex)/dy = ex
(ex)' = ex
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Proof of Derivative of ex
The derivative of cot x can be proved using the following ways:
- By using First Principle of Derivative
- By using Derivative of ax
Differentiation of ex Using First Principle
We are going to prove the differentiation of ex is ex by using the first principle of derivatives. We will use some basic rules and formulas of exponential functions and derivatives that are given below.
f'(x) = lim h→0 [f(x + h) - f(x)] / h
Also, we know that,
ex+h = ex.eh
lim x→0 (ex - 1) / x = 1
Using above formulas, we get,
d(ex)/dx = lim h→0 [ex+h - ex] / h
= lim h→0 [ex.eh - ex] / h
= lim h→0 ex [eh - 1] / h
= ex lim h→0 [eh - 1] / h
= ex × 1 = ex
Hence we have proved that the derivative of ex to be equal to ex
Differentiation of ex Using Derivative of ax
Exponential function are expressed in the form f(x) = ax , where 'a' is a constant (real number) and x is variable.
Derivative of exponential function f(x) = ax is find by the formula,
f'(x) = (ln a).ax...(i)
Substituting value a = e in eq (i), we get the differentiation of ex which is given by
f'(x) = (ln e) e x
= 1 × ex [As, ln e = 1]
= ex
Hence, the derivative of e to the power x is e x .
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Solved Examples on Derivative of ex
Some examples related to Derivative of ex are,
Example 1: Find the derivative of e2x.
Solution:
Given y = e2x
Using Chain Rule
⇒ y' = d(e2x)/dx
⇒ y' = [d(e2x)/dx ].[ d(2x)/dx]
⇒ y' = e2x. 2
⇒ y' = 2(e2x)
Example 2: Find the derivative of ex/x2.
Solution:
Given: y = ex/x2
Using Quotient Rule i.e.,
[(d[u/v]/dx) = [u’v – uv’]/v2]⇒ y' = d(ex/x2)/dx
⇒ y' = ( d(ex)/dx . x2 - ex. d(x2)/dx )/ (x2)2
⇒ y' = ( ex. x2 - ex. 2x )/ x4
⇒ y'= ( x .ex( x - 2))/x4
⇒ y' = ( ex. (x-2) )/ x3
Example 3: Find the derivative of the function (ex)x
Solution:
y= (ex)x
Taking log both side, we get,
ln y = ln (ex)x
⇒ ln y = x.(ln ex )
⇒ ln y = x2Differentiation both side, we get,
1/y . y' = 2x
⇒ y' = y. (2x)
⇒ y' = (ex)x. (2x)
Example 4: Evaluate the derivative of e2x+ x.sinx
Solution:
y = e2x + x.sinx
⇒ y' = d(e2x)/dx + d(x.sinx)/dxTo solve this we need to apply Chain rule for e2x and Product rule for x.sinx
y' = (d(e2x)/dx . d(2x)/dx ) + (dx/dy. sinx + x . d(sinx)/dy)
⇒ y' = e2x.2 + (1.sinx + x.cosx)
⇒ y' = 2.e2x + sinx + xcosx
Practice Questions on Derivative of ex
1. Find the derivative of e5x
2. Find the derivative of x3.e3x
3. Evaluate: (d/dx) [ex/(x2 + 2)]
4. Evaluate the derivative of: ex. log x