Derivative of e^-2x

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 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 (e^x-1)/x is 1 when x→0.

= -2e^-2x

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

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