Differentiation of Different Functions Revision Notes for VCE SSCE Mathematical Methods

Differentiation of Different Functions

Understanding exponential functions and their derivatives

When working with exponential functions, we discover that one particular base has remarkable properties that make it especially important in calculus. This special number is Euler's number, denoted as $e$ .

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The exponential function with base $e$ has a unique property that sets it apart from all other exponential functions - it is its own derivative. This remarkable characteristic makes it fundamental to calculus and appears throughout mathematics, physics, and engineering.

Investigating the derivative of exponential functions

To understand why $e$ is special, let's explore what happens when we differentiate exponential functions with different bases.

Consider the function $f(x) = 2^x$ . Using first principles to find its derivative:

f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

= \lim_{h \to 0} \frac{2^{x+h} - 2^x}{h}

= 2^x \lim_{h \to 0} \frac{2^h - 1}{h}

= 2^x f'(0)

Through numerical investigation, we find that f'(0) ≈ 0.6931, which means $f'(x) \approx 0.6931 \times 2^x$ .

Similarly, for $g(x) = 3^x$ , we can show that $g'(x) = 3^x g'(0)$ , where g'(0) ≈ 1.0986.

This leads us to an important question: can we find a number $b$ between $2$ and $3$ such that if $f(x) = b^x$ , then $f'(0) = 1$ and therefore $f'(x) = b^x$ ?

Finding Euler's number

Through systematic numerical investigation, we can narrow down this special value. The investigation involves calculating $\frac{b^h - 1}{h}$ as $h$ approaches $0$ , for various values of $b$ .

This investigation reveals that the special number $b$ is Euler's number $e \approx 2.71828$ . This gives us our first important result:

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For $f(x) = e^x$ , $f'(x) = e^x$

This means the exponential function with base $e$ is its own derivative - a truly remarkable property that makes it fundamental in calculus.

Differentiating functions involving $e^x$

The derivative of $e^{kx}$

When we have an exponential function of the form $y = e^{kx}$ where $k$ is a constant, we use the chain rule to find the derivative.

Let $u = kx$ . Then $y = e^u$ .

Using the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

= e^u \cdot k

= ke^{kx}

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For $f(x) = e^{kx}$ , $f'(x) = ke^{kx}$ where $k \in \mathbb{R}$

The graph illustrates this concept with $k = 2$ . Notice that:

The gradient of $y = e^x$ at point $P(1, e)$ is $e$
The gradient of $y = e^{2x}$ at point $Q(1, e^2)$ is $2e^2$

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Worked Example: Basic Exponential Derivatives

Find the derivative of each function with respect to $x$ :

a) $e^{3x}$

Let $y = e^{3x}$

\frac{dy}{dx} = 3e^{3x}

b) $e^{-2x}$

Let $y = e^{-2x}$

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

c) $e^{2x+1}$

We can rewrite this as $y = e^{2x} \cdot e = e \cdot e^{2x}$

\frac{dy}{dx} = 2e \cdot e^{2x} = 2e^{2x+1}

d) $\frac{1}{e^{2x}} + e^{3x}$

First, rewrite: $y = e^{-2x} + e^{3x}$

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

Using the chain rule with composite exponential functions

For more complex exponential functions where the exponent is itself a function, we apply the chain rule.

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For $h(x) = e^{f(x)}$ , the chain rule gives $h'(x) = f'(x)e^{f(x)}$

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Worked Example: Composite Exponential Functions

Find the derivative of each function with respect to $x$ :

a) $e^{x^2}$

Let $y = e^{x^2}$ and $u = x^2$

Then $y = e^u$

Using the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

= e^u \cdot 2x

= 2xe^{x^2}

b) $e^{x^2+4x}$

Let $y = e^{x^2+4x}$ and $u = x^2 + 4x$

Then $y = e^u$

Using the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

= e^u(2x + 4)

= (2x + 4)e^{x^2+4x}

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Worked Example: Finding Gradients of Tangent Lines

Find the gradient of the tangent to the curve $y = e^{2x} + 4$ at the point:

a) $(0, 5)$

First, find the derivative: $\frac{dy}{dx} = 2e^{2x}$

When $x = 0$ :

\frac{dy}{dx} = 2e^0 = 2

The gradient at $(0, 5)$ is 2.

b) $(1, e^2 + 4)$

Using $\frac{dy}{dx} = 2e^{2x}$

When $x = 1$ :

\frac{dy}{dx} = 2e^2

The gradient at $(1, e^2 + 4)$ is 2e².

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Worked Example: Derivatives with Given Function Values

For each expression, find the derivative with respect to $x$ , then evaluate at $x = 2$ , given that $f(2) = 0$ , $f'(2) = 4$ , and $f'(e^2) = 5$ .

a) $e^{f(x)}$

Let $y = e^{f(x)}$ and $u = f(x)$ . Then $y = e^u$ .

Using the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

= e^u f'(x)

= e^{f(x)} f'(x)

When $x = 2$ :

\frac{dy}{dx} = e^0 \times 4 = 4

b) $f(e^x)$

Let $y = f(e^x)$ and $u = e^x$ . Then $y = f(u)$ .

Using the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

= f'(u) \cdot e^x

= f'(e^x) \cdot e^x

When $x = 2$ :

\frac{dy}{dx} = f'(e^2) \cdot e^2 = 5e^2

Differentiation of the natural logarithm function

The natural logarithm function, written as $\log_e(x)$ or $\ln(x)$ , is the inverse of the exponential function $e^x$ . This relationship helps us find its derivative.

Deriving the derivative of $\log_e(kx)$

Let $y = \log_e(kx)$ where $k > 0$ , and solve for $x$ :

e^y = kx

x = \frac{1}{k}e^y

Since we know that $\frac{dx}{dy} = \frac{1}{k}e^y$ and $e^y = kx$ :

\frac{dx}{dy} = \frac{kx}{k} = x

Therefore:

\frac{dy}{dx} = \frac{1}{x}

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For $f(x) = \log_e(kx)$ where $k > 0$ , $f'(x) = \frac{1}{x}$

Notice that the constant $k$ doesn't affect the derivative - the result is simply $\frac{1}{x}$ .

The derivative of $\log_e(ax + b)$

For linear expressions inside the logarithm, we use the chain rule.

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If $y = \log_e(ax + b)$ for $x > -\frac{b}{a}$ , then $\frac{dy}{dx} = \frac{a}{ax + b}$

This is a particularly useful result that appears frequently in differentiation problems.

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Worked Example: Differentiating Logarithmic Functions

Find the derivative of each function with respect to $x$ :

a) $\log_e(5x)$ , $x > 0$

Method 1: Direct application

Let $y = \log_e(5x)$ for $x > 0$

\frac{dy}{dx} = \frac{1}{x}

Method 2: Using the chain rule

Let $u = 5x$ . Then $y = \log_e u$

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

= \frac{1}{u} \times 5

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

b) $\log_e(5x + 3)$ , $x > -\frac{3}{5}$

Let $y = \log_e(5x + 3)$ and $u = 5x + 3$

Then $y = \log_e u$

Using the chain rule:

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

= \frac{1}{u} \times 5

= \frac{5}{5x + 3}

Using the chain rule with composite logarithmic functions

When the logarithm contains a more complex function, we apply the chain rule systematically.

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For $h(x) = \log_e(f(x))$ , the chain rule gives $h'(x) = \frac{f'(x)}{f(x)}$

This means: differentiate what's inside and divide by what's inside.

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Worked Example: Composite Logarithmic Functions

Differentiate each function with respect to $x$ :

a) $\log_e(x^2 + 2)$

Let $y = \log_e(x^2 + 2)$ and $u = x^2 + 2$

Then $y = \log_e u$

Using the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

= \frac{1}{u} \cdot 2x

= \frac{2x}{x^2 + 2}

b) $(\log_e x)^2$ , $x > 0$

Let $y = (\log_e x)^2$ and $u = \log_e x$

Then $y = u^2$

Using the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

= 2u \cdot \frac{1}{x}

= \frac{2\log_e x}{x}

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Important Note About Logarithms of Negative Arguments

For $y = \log_e(-x)$ where $x < 0$ , using the chain rule with $u = -x$ gives:

\frac{dy}{dx} = \frac{1}{-x} \times (-1) = \frac{1}{x}

This shows that the derivative is still $\frac{1}{x}$ , even when the argument is negative (as long as the argument itself is positive).

Key formulas summary

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Key Formulas for Exponential and Logarithmic Differentiation

Exponential functions:

For exponential functions involving $e$ :

$\frac{d}{dx}(e^x) = e^x$
$\frac{d}{dx}(e^{kx}) = ke^{kx}$ where $k \in \mathbb{R}$
$\frac{d}{dx}(e^{f(x)}) = f'(x)e^{f(x)}$ (using the chain rule)

Natural logarithm functions:

For natural logarithm functions:

$\frac{d}{dx}(\log_e(kx)) = \frac{1}{x}$ where $k > 0$
$\frac{d}{dx}(\log_e(ax + b)) = \frac{a}{ax + b}$ where $x > -\frac{b}{a}$
$\frac{d}{dx}(\log_e(f(x))) = \frac{f'(x)}{f(x)}$ (using the chain rule)

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Key Points to Remember:

The exponential function $e^x$ is unique because it is its own derivative: $\frac{d}{dx}(e^x) = e^x$
When differentiating $e^{kx}$ , multiply by the coefficient $k$ to get $ke^{kx}$
For composite exponential functions $e^{f(x)}$ , use the chain rule: the derivative is $f'(x)e^{f(x)}$
The derivative of $\log_e(kx)$ is simply $\frac{1}{x}$ , regardless of the constant $k$ (provided $k > 0$ )
For composite logarithmic functions $\log_e(f(x))$ , the derivative is $\frac{f'(x)}{f(x)}$ - differentiate the inside and divide by the inside
Always check the domain when working with logarithms - the argument must be positive

Differentiation of Different Functions (VCE SSCE Mathematical Methods): Revision Notes