Derivative of e^2x: Formula, Proof by First Principle, Chain Rule

What is the Derivative of e2x? The derivative of e2x is 2e2x. Here, we will find the derivative of e2x using the derivative of ex. We will use three methods: substitution method, chain rule, and limit definition.

Derivative of e2x Formula

The derivative of e2x is 2e2x. This can be written mathematically as follows: d/dx(e2x) = 2 e2x  or (e2x)’ = 2 e2x.

What is the derivative of e2x?

At first, we will find the derivative of e2x by the substitution method. This method is known as logarithmic differentiation. The following steps have to be followed in this method.

Step 1: Let

\[y=e^{2x}.\]

Step 2: Taking logarithms on both sides, we get that

\[\log_e y =\log_e e^{2x}\]

\[\Rightarrow \log_e y =2x\log_e e \]

\[\Rightarrow \log_e y =2x [\because \log_a a=1]\]

Step 3: Differentiating both sides of logey=2x with respect to x, we have

$\dfrac{1}{y} \dfrac{dy}{dx}=\dfrac{d}{dx}(2x)$

$\Rightarrow \dfrac{1}{y} \dfrac{dy}{dx}=2$

$\Rightarrow \dfrac{dy}{dx}=2y$

$\Rightarrow \dfrac{dy}{dx}=2e^{2x}$ $[\because y=e^{2x}]$

∴ The derivative of e2x is 2e2x. That is, we have

\[\dfrac{d}{dx}(e^{2x})=2e^{2x}.\]

Derivative of e2x using chain rule

Recall the chain rule of derivatives: $\frac{df}{dx}=\frac{df}{dz} \cdot \frac{dz}{dx}.$ To find the differentiation of e2x by the chain rule, we put

z=2x

Now, $\dfrac{d}{dx}(e^{2x})=\dfrac{d}{dx}(e^z)$

$=\dfrac{d}{dz}(e^z) \cdot \dfrac{dz}{dx}$, by the chain rule.

$=\dfrac{d}{dz}(e^z) \cdot \dfrac{d}{dx}(2x)$

= ez ⋅ 2 as (ex)$’$=ex

= 2e2x  [ z=2x]

∴ the value of the derivative of e2x is 2e2x.

Also Read: 

 

Derivative of cube root of x: The derivative of the cube root of x is 1/(3x^{2/3})

Derivative of root x: The derivative of √x is 1/2√x

Integration of root x: The integration of  √x is 2/3x^{3/2}

Integration of mod x: The integration of  |x| is -x|x|/2 +c

Derivative of e2x from first principle

From first principle of derivatives, we know the derivative of f(x) is given by

$f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$

So taking f(x)=e2x, the derivative of e2x by first principle is

$(e^{2x})’= \lim\limits_{h \to 0} \dfrac{e^{2(x+h)}-e^{2x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{2x+2h}-e^{2x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{2x} \cdot e^{2h}-e^{2x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{2x}(e^{2h}-1)}{h}$

$=e^{2x}\lim\limits_{h \to 0} \Big(\dfrac{e^{2h}-1}{2h} \times 2 \Big)$

$=2e^{2x}\lim\limits_{h \to 0} \dfrac{e^{2h}-1}{2h}$

[Let $t=2h.$ Then $t \to 0$ as $x \to 0$]

$=2e^{2x}\lim\limits_{t \to 0} \dfrac{e^{t}-1}{t}$

$=2e^{2x} \cdot 1$

$=2e^{2x}$

This shows that the differentiation of e2x is 2e2x.

n-th Derivative of e2x

Differentiating e2x with respect to x for n times, we will get the nth derivative of e to the power x. The first order derivative of e2x is 2e2x. So the second order derivative of e2x is equal to d/dx(2e2x) = 2 d/dx(e2x) = 2×2 e2x = 22 e2x. Now, we will understant the patterns of these derivatives:

  • First derivative of e2x is 2e2x = 21 e2x
  • Second derivative of e2x is 2e2x
  • Third derivative of e2x is 23 e2x  and so on.

Observing these patterns carefully, we get the nth derivative of e2x is 2n e2x.

Important Notes on Derivative of e2x:

  • The differentiation of ex is ex but d/dx(e2x) = 2e2x.
  • The derivative of e-2x is -2e-2x.
  • In general, the derivative of emx is memx where m is a real number.

Application of Derivative of e2x

From the above we know that

$\dfrac{d}{dx}(e^{2x})$ $=2e^{2x}$ $\cdots (I)$

Using this fact, we can calculate the derivatives of many exponential functions. We will find the derivative of $e^{2x+1}.$

Question 1: Find $\dfrac{d}{dx}(e^{2x+1})$

Solution:

Let z=2x+1.

$\dfrac{d}{dx}(e^{2x+1})$ $=\dfrac{d}{dx}(e^z)$

$=\dfrac{d}{dz}(e^z) \cdot \dfrac{dz}{dx}$ by the chain rule

$=e^z \cdot \dfrac{d}{dx}(2x+1)$ by Equation (I)

$=e^z \cdot 2$

$=2e^{2x+1}$  $[\because z=2x+1]$

Thus, the derivative of e2x+1 is 2e2x+1.

Question 2: What is the derivative of e2?

Solution:

Note that the function eis a constant function of x. We know that the derivative of a constant function is 0. Thus, the derivative of e square is zero.

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