Note that e^{3x} is an exponential function. The derivative of e^{3x }is 3e^{3x}. To find the derivative of e^{3x}, we will use the below methods:

- Logarithmic differentiation
- Chain rule of derivatives
- First principle of derivatives

Table of Contents

## Derivative of e^{3x }Formula

The derivative of e^{3x }is 3e^{3x}. Mathematically, we can express it as

d/dx(e^{3x}) = 3e^{3x } or (e^{3x})’ = 3e^{3x}.

## What is the derivative of e^{3x}?

**Answer:** The derivative of e^{3x }is 3e^{3x}.

We will use the logarithmic differentiation to find the derivative of e^{3x}. Let us assume that

y = e^{3x}

Taking logarithms with base e to both sides, we obtain that

log_{e} y = log_{e} e^{3x}

⇒ log_{e} y = 3x as we know that log_{e} e^{a} = a.

Differentiating with respect to x, we get that

$\dfrac{1}{y} \dfrac{dy}{dx}=3$

⇒ $\dfrac{dy}{dx}=3y$

⇒ $\dfrac{dy}{dx}=3e^{3x}$

Thus, the derivative of e to the power 3x is 3e^{3x} and this is obtained by the logarithmic differentiation.

## Derivative of e^{3x} by First Principle

Let f(x)=e^{3x}. Using the limit definition of derivative or using the first principle of derivatives, the derivative of f(x) = e^{3x} is equal to

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

∴ $\dfrac{d}{dx}(e^{3x})= \lim\limits_{h \to 0} \dfrac{e^{3(x+h)}-e^{3x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{3x+3h}-e^{3x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{3x} \cdot e^{3h}-e^{3x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{3x}(e^{3h}-1)}{h}$

=e^{3x} $\lim\limits_{h \to 0} \Big(\dfrac{e^{3h}-1}{3h} \times 3 \Big)$

= 3e^{3x} $\lim\limits_{h \to 0} \dfrac{e^{3h}-1}{3h}$

[Let t=3h. Then t→0 as x →0]

= 3e^{3x} $\lim\limits_{t \to 0} \dfrac{e^{t}-1}{t}$

= 3e^{3x} ⋅ 1

= 3e^{3x}

∴ The differentiation of e^{3x} is 3e^{3x} and this is achieved from the first principle of derivatives.

## Derivative of e^{3x} by Chain Rule

Note that the exponential function e^{3x} can be written as a composite function in the following way:

f(g(x)) = e^{3x},

where f(x)=e^{x} and g(x)=3x.

⇒ $f'(x)=e^x$ and $g'(x)=3$.

By the chain rule, the derivative of f(g(x)) is equal to f'(g(x)) g'(x).

∴ The differentiation of e^{3x} by chain rule is equal to

[f(g(x))]$’$ = f$’$(g(x)) g$’$(x)

⇒ (e^{3x})$’$= f$’$(3x) ⋅ 3

= e^{3x} ⋅ 3

= 3e^{3x}

∴ the value of the derivative of e^{3x }is 3e^{3x}.