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

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.

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