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

log_{e} y = log_{e} e^{-3x}

⇒ log_{e} y = -3x by the logarithm rule log_{e} e^{a} = a.

Differentiating both sides with respect to x, we get that

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

⇒ $\dfrac{dy}{dx}=-3y$

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

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

**Also Read:**

Derivative of log 3x: The derivative of log(3x) is 1/x.Integration of root x: The integration of root(x) is 2/3x^{3/2}+c.Derivative of 1/x: The derivative of 1 by x is -1/x^{2}. |

## 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.

d/dx(e^{-3x}) | = d/du(e^{u}) × d/dx(-3x) |

= e^{u} × -3 | |

= -3e^{u} | |

= -3e^{-3x} as 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 (e^{x}-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 e**

^{3x}+e^{-3x}?Answer: The derivative of e^{3x}+e^{-3x }is 3(e^{3x}-e^{-3x}).