Derivative of e^3x: Formula, Proof by First Principle

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

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.