Differentiation of Exponential Functions Revision Notes for HSC SSCE Mathematics Advanced

Differentiation of Exponential Functions

Introduction

Building on previous knowledge of exponential functions, we now explore how to differentiate functions involving $e^x$ . Two fundamental standard forms provide the foundation for this work:

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

$\frac{d}{dx}e^{ax+b} = ae^{ax+b}$

These forms allow us to differentiate exponential functions directly and form the basis for more complex differentiation techniques.

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The exponential function $e^x$ has the unique property that its derivative equals itself. This is a remarkable characteristic that makes exponential functions particularly important in calculus and its applications.

Using the two standard forms

The standard forms can be applied directly to many exponential functions. The first form shows that the exponential function $e^x$ has the unique property that its derivative equals itself. The second form applies when the exponent is a linear expression $ax + b$ , where the derivative includes the coefficient $a$ as a multiplier.

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Worked Example: Basic differentiation

Let's differentiate the following functions using the standard forms:

Part a: $y = e^x + e^{-x}$

Given $y = e^x + e^{-x}$

For $e^{-x}$ , we have $a = -1$ and $b = 0$

$y' = e^x - e^{-x}$

Part b: $y = 5e^{4x-3}$

Given $y = 5e^{4x-3}$

Here $a = 4$ and $b = -3$

$y' = 5 \times 4e^{4x-3}$

$y' = 20e^{4x-3}$

Part c: $y = e^{2-\frac{1}{2}x}$

Given $y = e^{2-\frac{1}{2}x}$

Here $a = -\frac{1}{2}$ and $b = 2$

$y' = -\frac{1}{2}e^{2-\frac{1}{2}x}$

Part d: $y = \sqrt{e^x} + \frac{1}{\sqrt{e^x}}$

First, rewrite using exponential notation:

$y = e^{\frac{1}{2}x} + e^{-\frac{1}{2}x}$

$y' = \frac{1}{2}e^{\frac{1}{2}x} - \frac{1}{2}e^{-\frac{1}{2}x}$

Differentiating using the chain rule

When the exponent is more complex than a linear expression, we use the chain rule. This powerful technique allows us to differentiate composite functions by breaking them into simpler parts.

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The Chain Rule

The chain rule states:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

This is essential when the exponent contains anything more complex than a linear expression (such as $x^2$ , trigonometric functions, or other composite expressions).

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Worked Example: Chain rule applications

Part a: $y = e^{1-x^2}$

Let $u = 1 - x^2$

Then $y = e^u$

Hence $\frac{du}{dx} = -2x$

and $\frac{dy}{du} = e^u$

Applying the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

$\frac{dy}{dx} = e^u \times (-2x)$

$\frac{dy}{dx} = -2xe^{1-x^2}$

Part b: $y = (e^{2x} - 3)^4$

Let $u = e^{2x} - 3$

Then $y = u^4$

Hence $\frac{du}{dx} = 2e^{2x}$

and $\frac{dy}{du} = 4u^3$

Applying the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

$\frac{dy}{dx} = 4u^3 \times 2e^{2x}$

$\frac{dy}{dx} = 4(e^{2x} - 3)^3 \times 2e^{2x}$

$\frac{dy}{dx} = 8e^{2x}(e^{2x} - 3)^3$

Part c: Proving the standard form $y = e^{ax+b}$

Let $u = ax + b$

Then $y = e^u$

Hence $\frac{du}{dx} = a$

and $\frac{dy}{du} = e^u$

Applying the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

$\frac{dy}{dx} = e^u \times a$

$\frac{dy}{dx} = ae^{ax+b}$

This proves the second standard form.

Alternative formula for the chain rule

Some students prefer to use a formula version of the chain rule. This can be written in two equivalent ways:

$\frac{d}{dx}e^u = e^u\frac{du}{dx} \quad \text{OR} \quad \frac{d}{dx}e^{f(x)} = e^{f(x)}f'(x)$

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While not essential to memorise, these formulas can provide a quicker method once you're confident with the chain rule concept. The key is to always show your working clearly.

Similarly, for differentiating powers of functions:

$\frac{d}{dx}u^n = nu^{n-1}\frac{du}{dx} \quad \text{OR} \quad \frac{d}{dx}(f(x))^n = n(f(x))^{n-1}f'(x)$

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Worked Example: Using the formula approach

Part a: $y = e^{1-x^2}$

Using the $u$ notation:

Let $u = 1 - x^2$

Then $\frac{du}{dx} = -2x$

Using $\frac{d}{dx}e^u = e^u\frac{du}{dx}$ :

$y' = e^{1-x^2} \times (-2x)$

$y' = -2xe^{1-x^2}$

Alternatively, using the $f(x)$ notation:

Let $f(x) = 1 - x^2$

Then $f'(x) = -2x$

Using $\frac{d}{dx}e^{f(x)} = e^{f(x)}f'(x)$ :

$y' = e^{1-x^2} \times (-2x)$

$y' = -2xe^{1-x^2}$

Part b: $y = (e^{2x} - 3)^4$

Let $u = e^{2x} - 3$

Then $\frac{du}{dx} = 2e^{2x}$

Using $\frac{d}{dx}u^4 = 4u^3\frac{du}{dx}$ :

$y' = 4(e^{2x} - 3)^3 \times 2e^{2x}$

$y' = 8e^{2x} \times (e^{2x} - 3)^3$

Three standard derivatives for exponential functions

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Essential Formulas for Exponential Differentiation

The following three formulas are essential for differentiating exponential functions:

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

$\frac{d}{dx}e^{ax+b} = ae^{ax+b}$

$\frac{d}{dx}e^u = e^u\frac{du}{dx} \quad \text{OR} \quad \frac{d}{dx}e^{f(x)} = e^{f(x)}f'(x)$

Using the product rule

When an exponential function is multiplied by another function, we use the product rule. For a function $y = uv$ where both $u$ and $v$ are functions of $x$ :

$\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$

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The result can often be factorised to help identify stationary points. Always look for common factors, particularly exponential terms and powers of $x$ .

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Worked Example: Product rule with exponential functions

Part a: $y = x^3e^x$

Let $u = x^3$ and $v = e^x$

Then $\frac{du}{dx} = 3x^2$ and $\frac{dv}{dx} = e^x$

Applying the product rule:

$\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$

$\frac{dy}{dx} = e^x \times 3x^2 + x^3 \times e^x$

Taking out the common factor $x^2e^x$ :

$\frac{dy}{dx} = x^2e^x(3 + x)$

For stationary points, set $\frac{dy}{dx} = 0$ :

$x^2e^x(3 + x) = 0$

Since $e^x$ is never zero, we have $x^2 = 0$ or $3 + x = 0$

Therefore $x = 0$ or $x = -3$

When $x = 0$ : $y = 0^3e^0 = 0$

When $x = -3$ : $y = (-3)^3e^{-3} = -27e^{-3}$

The stationary points are (0, 0) and (-3, -27e^{-3})

Part b: $y = xe^{5x-2}$

Let $u = x$ and $v = e^{5x-2}$

Then $u' = 1$ and $v' = 5e^{5x-2}$

Applying the product rule:

$y' = vu' + uv'$

$y' = e^{5x-2} \times 1 + x \times 5e^{5x-2}$

Taking out the common factor $e^{5x-2}$ :

$y' = e^{5x-2}(1 + 5x)$

For stationary points, set $y' = 0$ :

$e^{5x-2}(1 + 5x) = 0$

Since $e^{5x-2}$ is never zero:

$1 + 5x = 0$

$x = -\frac{1}{5}$

When $x = -\frac{1}{5}$ : $y = -\frac{1}{5}e^{5(-\frac{1}{5})-2} = -\frac{1}{5}e^{-3}$

The stationary point is (-\frac{1}{5}, -\frac{1}{5}e^{-3})

Using the quotient rule

When an exponential function appears in a fraction, we use the quotient rule. For a function $y = \frac{u}{v}$ where both $u$ and $v$ are functions of $x$ :

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

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Finding Stationary Points with the Quotient Rule

For a stationary point when using the quotient rule, the numerator must equal zero (while the denominator remains non-zero). This simplifies the process of finding stationary points.

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Worked Example: Quotient rule with exponential functions

Part a: $y = \frac{e^{5x}}{x}$

Let $u = e^{5x}$ and $v = x$

Then $\frac{du}{dx} = 5e^{5x}$ and $\frac{dv}{dx} = 1$

Applying the quotient rule:

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

$\frac{dy}{dx} = \frac{x \times 5e^{5x} - e^{5x} \times 1}{x^2}$

$\frac{dy}{dx} = \frac{5xe^{5x} - e^{5x}}{x^2}$

Taking out the common factor $e^{5x}$ in the numerator:

$\frac{dy}{dx} = \frac{e^{5x}(5x - 1)}{x^2}$

For a stationary point, the numerator must equal zero:

$e^{5x}(5x - 1) = 0$

Since $e^{5x}$ is never zero:

$5x - 1 = 0$

$x = \frac{1}{5}$

Part b: $y = \frac{e^x}{1 - x^2}$

Let $u = e^x$ and $v = 1 - x^2$

Then $u' = e^x$ and $v' = -2x$

Applying the quotient rule:

$y' = \frac{vu' - uv'}{v^2}$

$y' = \frac{(1 - x^2)e^x - e^x(-2x)}{(1 - x^2)^2}$

$y' = \frac{(1 - x^2)e^x + 2xe^x}{(1 - x^2)^2}$

Taking out the common factor $e^x$ in the numerator:

$y' = \frac{e^x(1 - x^2 + 2x)}{(1 - x^2)^2}$

$y' = \frac{e^x(1 + 2x - x^2)}{(1 - x^2)^2}$

For a stationary point, the numerator must equal zero:

$e^x(1 + 2x - x^2) = 0$

Since $e^x$ is never zero:

$1 + 2x - x^2 = 0$

Rearranging: $x^2 - 2x - 1 = 0$

Using the quadratic formula with $\triangle = 4 + 4 = 8$ :

$x = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$

Therefore x = 1 + \sqrt{2} or x = 1 - \sqrt{2}

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Key Points to Remember:

The derivative of $e^x$ is itself: $\frac{d}{dx}e^x = e^x$
For linear exponents, multiply by the coefficient: $\frac{d}{dx}e^{ax+b} = ae^{ax+b}$
Use the chain rule for composite exponential functions: $\frac{d}{dx}e^u = e^u\frac{du}{dx}$
Apply the product rule when exponential functions are multiplied by other functions
Apply the quotient rule when exponential functions appear in fractions
Factorise derivatives to find stationary points easily
Remember that e^x is never zero, which simplifies finding stationary points

Differentiation of Exponential Functions (HSC SSCE Mathematics Advanced): Revision Notes