Derivatives of Exponential Functions

The natural exponential $e^{x}$ is its own derivative, making it uniquely simple. For other bases, the derivative of $a^{x}$ picks up a factor of $\ln a$ .

Function	Derivative
$e^{x}$	$e^{x}$
$a^{x}$ ( $a > 0$ )	$(\ln a) a^{x}$
$e^{u}$ (chain)	$e^{u} \frac{d u}{d x}$
$a^{u}$ (chain)	$(\ln a) a^{u} \frac{d u}{d x}$
$u^{v}$ (both variable)	$\ln u \cdot u^{v} \frac{d v}{d x} + v \cdot u^{v - 1} \frac{d u}{d x}$

Derivative of $a^{x}$

Let $y = a^{x}$ ( $a > 0$ ). Taking $\ln$ of both sides:

\ln y = x \ln a \Rightarrow x = \frac{\ln y}{\ln a}

Differentiating with respect to $y$ :

\frac{d x}{d y} = \frac{1}{\ln a} \cdot \frac{1}{y}

By the Derivative Rule for Inverses, $\frac{d y}{d x} = \ln a \cdot y$ , so:

\boxed{\frac{d}{dx}a^x=(\ln a)\,a^x\tag{c}}

When $a = e$ , $\ln e = 1$ :

\boxed{\frac{d}{dx}e^x=e^x\tag{d}}

Derivative of $u^{v}$ When Both Are Variable

Let $y = u^{v}$ where $u = u (x) > 0$ and $v = v (x)$ . Write $y = e^{v \ln u}$ and apply (d) with the Chain Rule:

\begin{aligned} \frac{dy}{dx}&=e^{v\ln u}\frac{d}{dx}(v\ln u)\\ &=u^v\left[\frac{dv}{dx}\ln u+v\cdot\frac{1}{u}\frac{du}{dx}\right]\\ &=\ln u\cdot u^v\frac{dv}{dx}+v\cdot u^{v-1}\frac{du}{dx} \end{aligned}\boxed{\frac{d}{dx}u^v=\ln u\cdot u^v\frac{dv}{dx}+v\cdot u^{v-1}\frac{du}{dx}\tag{e}}

Interpretation: The derivative of $u^{v}$ is the sum of two terms: one obtained by differentiating while treating the base $u$ as constant, and another by treating the exponent $v$ as constant.

Worked Examples

If $y = a^{3 x^{2}}$ where $a$ is a constant, find $d y / d x$ .

Solution

**Method (a):** Let

v = 3 x^{2}

, use (c) and the Chain Rule:

\frac{d y}{d x} = (\ln a) a^{v} \cdot 6 x = 6 x \ln a \cdot a^{3 x^{2}}

**Method (b):** Logarithmic differentiation.

\ln y = 3 x^{2} \ln a

, so y'/y=6x\ln a, giving y'=6x\ln a\cdot a^{3x^2}.

Differentiate $y = b e^{c^{2} + x^{2}}$ with respect to $x$ .

Solution

Let

v = c^{2} + x^{2}

. By (d) and the Chain Rule:

\frac{d y}{d x} = b e^{v} \cdot \frac{d v}{d x} = b e^{c^{2} + x^{2}} \cdot 2 x = 2 b x e^{c^{2} + x^{2}}

Differentiate $y = x^{(e^{x})}$ .

Solution

**Method (a):** Use formula (e) with

u = x

and

v = e^{x}

\frac{d y}{d x} = \ln x \cdot x^{e^{x}} \cdot e^{x} + e^{x} \cdot x^{e^{x} - 1} \cdot 1 = e^{x} x^{e^{x}} (\ln x + \frac{1}{x})

**Method (b):** Logarithmic differentiation.

\ln y = e^{x} \ln x

, so \frac{y'}{y}=e^x\ln x+e^x\cdot\frac{1}{x}=e^x\!\left(\ln x+\frac{1}{x}\right) y'=x^{e^x}\cdot e^x\!\left(\ln x+\frac{1}{x}\right)

Find $d y / d x$ if $y = {(\frac{x^{2}}{\sqrt{x^{2} + 1}})}^{\sin x}$ .

Solution

Let

y = u^{v}

with

u = x^{2} / \sqrt{x^{2} + 1}

and

v = \sin x

. Using logarithmic differentiation:

\ln y = \sin x \cdot \ln \frac{x^{2}}{\sqrt{x^{2} + 1}} = \sin x [2 \ln x - \frac{1}{2} \ln (x^{2} + 1)]

Differentiating: \frac{y'}{y}=\cos x\left[2\ln x-\frac{1}{2}\ln(x^2+1)\right]+\sin x\left[\frac{2}{x}-\frac{x}{x^2+1}\right] y'=\left(\frac{x^2}{\sqrt{x^2+1}}\right)^{\sin x}\left\{\cos x\cdot\ln\frac{x^2}{\sqrt{x^2+1}}+\sin x\left[\frac{2}{x}-\frac{x}{x^2+1}\right]\right\}

Frequently Asked Questions

Why is

e^{x}

its own derivative?

This follows from the definition of

e

as the base for which

\ln e = 1

. In formula (c),

\frac{d}{d x} e^{x} = (\ln e) e^{x} = e^{x}

What is

\frac{d}{d x} 2^{x}

\frac{d}{d x} 2^{x} = (\ln 2) 2^{x}

. Since

\ln 2 \approx 0.693

, the derivative is less than the function itself (unlike

e^{x}

where they are equal).

What is the difference between

x^{e}

and

e^{x}

x^{e}

has a constant exponent, so

\frac{d}{d x} x^{e} = e x^{e - 1}

(Power Rule).

e^{x}

has a constant base and variable exponent, so

\frac{d}{d x} e^{x} = e^{x}

. When both base and exponent vary, use formula (e) or logarithmic differentiation.

Derivative of $a^{x}$

Derivative of $u^{v}$ When Both Are Variable

Worked Examples

Frequently Asked Questions

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Derivatives of Exponential Functions

Derivative of a x

Derivative of u v When Both Are Variable

Worked Examples

Frequently Asked Questions

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Derivative of $a^{x}$

Derivative of $u^{v}$ When Both Are Variable