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

Note that e3x is an exponential function. The derivative of e3x is 3e3x. To find the derivative of e3x, we will use the below methods:

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

Derivative of e3x Formula

The derivative of e3x is 3e3x. Mathematically, we can express it as

d/dx(e3x) = 3e3x  or (e3x)’ = 3e3x.

What is the derivative of e3x?

Answer: The derivative of e3x is 3e3x.

We will use the logarithmic differentiation to find the derivative of e3x. Let us assume that

y = e3x

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

loge y = loge e3x

⇒ loge y = 3x as we know that loge ea = 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 3e3x and this is obtained by the logarithmic differentiation.

Derivative of e3x by First Principle

Let f(x)=e3x. Using the limit definition of derivative or using the first principle of derivatives, the derivative of f(x) = e3x is equal to

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

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

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

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

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

=e3x $\lim\limits_{h \to 0} \Big(\dfrac{e^{3h}-1}{3h} \times 3 \Big)$

= 3e3x $\lim\limits_{h \to 0} \dfrac{e^{3h}-1}{3h}$

[Let t=3h. Then t→0 as x →0]

= 3e3x $\lim\limits_{t \to 0} \dfrac{e^{t}-1}{t}$

= 3e3x ⋅ 1

= 3e3x

∴ The differentiation of e3x is 3e3x and this is achieved from the first principle of derivatives.

Derivative of e3x by Chain Rule

Note that the exponential function e3x can be written as a composite function in the following way:

f(g(x)) = e3x,

where f(x)=ex 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 e3x by chain rule is equal to

[f(g(x))]$’$ = f$’$(g(x)) g$’$(x)

⇒ (e3x)$’$= f$’$(3x) ⋅ 3

= e3x ⋅ 3

= 3e3x

∴ the value of the derivative of e3x is 3e3x.

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