# Derivative of e^-x: Proof by First Principle, Chain Rule

The function e to the power -x is an exponential function, denoted by e-x. The derivative of e-x is equal to -e-x. In this post, we will learn how to find the derivative of e-x by different methods.

## Derivative of e-x Formula

The derivative of e-x is -e-x. Mathematically, this can be expressed as follows:

d/dx(e-x) = -e-x  or (e-x)’ = -e-x.

This will be proved here using the following methods:

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

## What is the derivative of e-x?

Answer: The derivative of e to the power -x is -e-x.

Proof: Let us use the logarithmic differentiation to find the derivative of e-x. We put

y = e-x

Taking logarithms with base e, we obtain that

loge y = loge e-x

⇒ loge y = -x by the logarithm rule loge ea = a.

Differentiating both sides with respect to x, we get that

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

⇒ $\dfrac{dy}{dx}=-y$

⇒ $\dfrac{dy}{dx}=-e^{-x}$

Thus, the derivative of e to the power -x is -e-x and this is obtained by the logarithmic differentiation.

## Derivative of e-x by First Principle

By the first principle of derivatives, the derivative of f(x) is equal to

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

Let f(x)=e-x.

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

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

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

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

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

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

[Let t=-h. Then t→0 as x →0]

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

= -e-x ⋅ 1 as the limit of (ex-1)/x is 1 when x→0.

= -e-x

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

## Derivative of e-x by Chain Rule

To find the derivative of a composite function, we use the chain rule. It says that the derivative of f(g(x)) is equal to

[f(g(x))]$’$ = f$’$(g(x)) g$’$(x) …(I)

The function e-x can be written as a composite function in the following way:

f(g(x)) = e-x,

where f(x)=ex and g(x)=-x.

⇒ $f'(x)=e^x$ and $g'(x)=-1$.

∴ By the above chain rule (I), the derivative of e-x is equal to

(e-x)$’$= f$’$(-x) ⋅ (-1)

= e-x ⋅ (-1)

= -e-x

∴ The value of the derivative of e-x by the chain rule is -e-x.

## FAQs on Derivative of e-x

Q1: What is the derivative of e-x?

Answer: The derivative of e-x is -e-x.

Q2: What is the derivative of ex+e-x?

Answer: The derivative of ex+e-x is ex-e-x.

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