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

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

⇒ log_{e} y = -x by the logarithm rule log_{e} e^{a} = 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 (e^{x}-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)=e^{x} 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 e**

^{x}+e^{-x}?Answer: The derivative of e^{x}+e^{-x }is e^{x}-e^{-x}.