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

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 2/3x3/2+c. Derivative of 1/x: The derivative of 1 by x is -1/x2.

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(eu) × d/dx(-3x)
	= eu × -3
	= -3eu
	= -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.

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