Derivative of e^4x: Formula, Proof by First Principle, Chain Rule

The derivative of e^4x is 4e^4x. Note that e^4x is an exponential function with exponential 4x. Here, we will find the derivative of e to the power 4x using the following rules:

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

Derivative of e^4x Formula

The derivative of e^4x is 4e^4x. Mathematically, we can write it as

d/dx(e^4x) = 4e^4x  or (e^4x)’ = 4e^4x.

What is the derivative of e^4x?

Answer: The derivative of e^4x is 4e^4x.

Proof: By the logarithmic differentiation, we will find the derivative of e^4x. Let us assume that

y = e^4x

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

log_e y = log_e e^4x

⇒ log_e y = 4x by the logarithm rule log_e e^a = a.

Differentiating with respect to x, we get that

$\dfrac{1}{y} \dfrac{dy}{dx}=4$

⇒ $\dfrac{dy}{dx}=4y$

⇒ $\dfrac{dy}{dx}=4e^{4x}$

This shows that the derivative of e^4x is 4e^4x and this is obtained by the logarithmic differentiation.

Also Read:

Derivative of esin x: The derivative of esin x is cos x esin x. Integration of modulus of x: The integration of mod x is -x|x|/2+c. Derivative of 1/x: The derivative of 1 by x is -1/x2.

Derivative of e^4x by Chain Rule

To find the derivative of a composite function, we use the chain rule. We will now find the derivative of e to the power 4x by the chain rule.

Let z=4x.

d/dx(e4x)	= d/dz(ez) × d/dx(4x)
	= ez × 4
	= 4ez
	= 4ez as z=4x.

Derivative of e^4x by First Principle

By the first principles, the derivative of a function f(x) is given by the following limit:

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

Put f(x)=e^4x.

So the derivative of e^4x by first principle is

$\dfrac{d}{dx}(e^{4x})= \lim\limits_{h \to 0} \dfrac{e^{4(x+h)}-e^{4x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{4x+4h}-e^{4x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{4x} \cdot e^{4h}-e^{4x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{4x}(e^{4h}-1)}{h}$

=e^4x $\lim\limits_{h \to 0} \Big(\dfrac{e^{4h}-1}{4h} \times 4 \Big)$

= 4e^4x $\lim\limits_{h \to 0} \dfrac{e^{4h}-1}{4h}$

[Let t=4h. Then t→0 as x →0]

= 4e^4x $\lim\limits_{t \to 0} \dfrac{e^{t}-1}{t}$

= 4e^4x ⋅ 1

= 4e^4x

∴ The differentiation of e^4x is 4e^4x and this is achieved from the first principle of derivatives.

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