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.
Table of Contents
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
loge y = loge e-2x
⇒ loge y = -2x by the logarithm rule loge ea = 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 esin x: The derivative of esin x is cos x esin 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/x2. |
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(eu) × d/dx(-2x) |
= eu × -2 | |
= -2eu | |
= -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 (ex-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
Answer: The derivative of e-2x is -2e-2x.
Answer: The derivative of e2x+e-2x is 2(e2x-e-2x).
Answer: The integration of e-2x is -e-2x/2+c.
This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.