# Derivative of e^-3x

The function e to the power -3x is written as e-3x and its derivative is -3e-3x. Here, we will find the derivative of e-3x by the following methods:

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

## Derivative of e-3x Formula

The derivative of e-3x is -3e-3x. Mathematically, we can express it as follows:

d/dx(e-3x) = -3e-3x  or (e-3x)’ = -3e-3x.

## What is the derivative of e-3x?

Answer: The derivative of e to the power -3x is -3e-3x.

Proof: Let us use the logarithmic differentiation to find the derivative of e-3x. We put

y = e-3x

Taking logarithms with base e, we obtain that

loge y = loge e-3x

⇒ loge y = -3x by the logarithm rule loge ea = a.

Differentiating both sides with respect to x, we get that

$\dfrac{1}{y} \dfrac{dy}{dx}=-3$

⇒ $\dfrac{dy}{dx}=-3y$

⇒ $\dfrac{dy}{dx}=-3e^{-3x}$ as y=e-3x.

Thus, the derivative of e to the power -3x is -3e-3x which is obtained by the logarithmic differentiation method.

## Derivative of e-3x by Chain Rule

To find the derivative of a composite function, we use the chain rule. Let us now find the derivative of e to the power -3x by the chain rule.

Let u=-3x.

Thus, the derivative of e-3x by the chain rule is -3e-3x.

## Derivative of e-3x by First Principle

Recall the first principle of derivatives: Let f(x) be a function of the variable x. The derivative of f(x) from the first principle is equal to the following limit:

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

We put f(x)=e-3x in the above formula. Then the derivative of e to the power -3x by the chain rule is given by

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

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

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

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

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

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

[Let t = -3h. Then t→0 as x →0]

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

= -3e-3x ⋅ 1 as the limit of (ex-1)/x is 1 when x→0.

= -3e-3x

∴ The differentiation of e-3x by the first principle is -3e-3x.

## FAQs on Derivative of e-3x

Q1: What is the derivative of e to the power -3x?

Answer: The derivative of e-3x is -3e-3x.

Q2: What is the integration of e-3x?

Answer: The integration of e-3x is -e-3x/3+c, where c is an integration constant.

Q3: What is the derivative of e3x+e-3x?

Answer: The derivative of e3x+e-3x is 3(e3x-e-3x).

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