The function e to the power -2x is an exponential function, denoted by e^{-2x}. The derivative of e^{-2x} is equal to -2e^{-2x}. In this post, we will learn how to find the derivative of e^{-2x} by the following methods:

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

## Derivative of e^{-2x }Formula

The derivative of e^{-2x }is -2e^{-2x}. Symbolically, we can express it as follows:

d/dx(e^{-2x}) = -2e^{-2x } or (e^{-2x})’ = -2e^{-2x}.

## What is the derivative of e^{-2x}?

**Answer:** The derivative of e to the power -2x is -2e^{-2x}.

*Proof:* Let us use the logarithmic differentiation to find the derivative of e^{-2x}. We put

y = e^{-2x}

Taking logarithms with base e, we obtain that

log_{e} y = log_{e} e^{-2x}

⇒ log_{e} y = -2x by the logarithm rule log_{e} e^{a} = a.

Differentiating both sides with respect to x, we get that

$\dfrac{1}{y} \dfrac{dy}{dx}=-2$

⇒ $\dfrac{dy}{dx}=-2y$

⇒ $\dfrac{dy}{dx}=-2e^{-2x}$ as y=e^{-2x}.

Thus, the derivative of e to the power -2x is -2e^{-2x} and this is obtained by the logarithmic differentiation method.

**Also Read:**

Derivative of e: The derivative of e^{sin x}^{sin x} is cos x e^{sin x}.Integration of modulus of x: The integration of mod x is -x|x|/2+c.Derivative of 1/x: The derivative of 1 by x is -1/x^{2}. |

## Derivative of e^{-2x} by Chain Rule

To find the derivative of a composite function, we use the chain rule. We will now find the derivative of e to the power -2x by the chain rule.

Let u=-2x.

d/dx(e^{-2x}) | = d/du(e^{u}) × d/dx(-2x) |

= e^{u} × -2 | |

= -2e^{u} | |

= -2e^{-2x} as u=-2x. |

Thus, the derivative of e^{-2x} by the chain rule is -2e^{-2x}.

## Derivative of e^{-2x} by First Principle

By the first principle of derivatives, the derivative of a function f(x) 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^{-2x} in the above formula. Then we obtain the derivative of e to the power -2x by the chain rule which is equal to

$\dfrac{d}{dx}(e^{-2x})= \lim\limits_{h \to 0} \dfrac{e^{-2(x+h)}-e^{-2x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{-2x-2h}-e^{-2x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{-2x} \cdot e^{-2h}-e^{-2x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{-2x}(e^{-2h}-1)}{h}$

=e^{-2x} $\lim\limits_{h \to 0} \Big(\dfrac{e^{-2h}-1}{-2h} \times (-2) \Big)$

= -2e^{-2x} $\lim\limits_{h \to 0} \dfrac{e^{-2h}-1}{-2h}$

[Let t = -2h. Then t→0 as x →0]

= -2e^{-2x} $\lim\limits_{t \to 0} \dfrac{e^{t}-1}{t}$

= -2e^{-2x} ⋅ 1 as the limit of (e^{x}-1)/x is 1 when x→0.

= -2e^{-2x}

∴ The differentiation of e^{-2x} by the first principle is -2e^{-2x}.

## FAQs on Derivative of e^{-2x}

**Q1: What is the derivative of e**

^{-2x}?Answer: The derivative of e^{-2x} is -2e^{-2x}.

**Q2: What is the derivative of e**

^{2x}+e^{-2x}?Answer: The derivative of e^{2x}+e^{-2x }is 2(e^{2x}-e^{-2x}).

**Q3: What is the integration of e**

^{-2x}?Answer: The integration of e^{-2x} is -e^{-2x}/2+c.