Derivatives and Their Applications

The Derivative as a Rate of Change

The derivative of $f(x)$ with respect to $x$ is written as $f'(x)$ or $\frac{dy}{dx}$. Geometrically, $f'(a)$ gives the gradient of the tangent line to the curve $y = f(x)$ at the point where $x = a$.

If $f'(a) > 0$, the function is increasing at $x = a$. If $f'(a) < 0$, the function is decreasing. If $f'(a) = 0$, the tangent is horizontal — this is a stationary point.

The Power Rule

The most fundamental differentiation rule:

$\text{If } f(x) = x^n, \text{ then } f'(x) = nx^{n-1}$

This works for any real exponent $n$, including negative and fractional values:

• $f(x) = x^3 \Rightarrow f'(x) = 3x^2$
• $f(x) = x^{-1} = \frac{1}{x} \Rightarrow f'(x) = -x^{-2} = -\frac{1}{x^2}$
• $f(x) = x^{1/2} = \sqrt{x} \Rightarrow f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

Constants differentiate to zero: if $f(x) = c$, then $f'(x) = 0$.

The Chain Rule

Used to differentiate composite functions $f(g(x))$:

$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

Informally: "differentiate the outer function, keep the inner function unchanged, then multiply by the derivative of the inner function."

Example: $y = (3x + 1)^5$

Let $u = 3x + 1$, so $y = u^5$.

$\frac{dy}{dx} = 5u^4 \cdot 3 = 15(3x+1)^4$

The Product Rule

Used when differentiating a product of two functions $y = u(x) \cdot v(x)$:

$\frac{d}{dx}[uv] = u'v + uv'$

Remember: differentiate each factor in turn while keeping the other unchanged, then add.

Example: $y = x^2 \sin x$

Let $u = x^2$ and $v = \sin x$.

$\frac{dy}{dx} = 2x \sin x + x^2 \cos x$

The Quotient Rule

Used when differentiating a quotient $y = \frac{u(x)}{v(x)}$:

$\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}$

This is in your formula booklet. A useful mnemonic: "low d-high minus high d-low, over the square of what's below."

Derivatives of Standard Functions

From your IB formula booklet (these must be recognised, not memorised from scratch):

$f(x)$	$f'(x)$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x = \frac{1}{\cos^2 x}$
$e^x$	$e^x$
$\ln x$	$\frac{1}{x}$
$a^x$	$a^x \ln a$

Tangent and Normal Lines

At a point $(a, f(a))$:

• The tangent line has gradient $m = f'(a)$ and equation $y - f(a) = f'(a)(x - a)$
• The normal line is perpendicular to the tangent, so its gradient is $m_n = -\frac{1}{f'(a)}$ (provided $f'(a) \neq 0$)

Stationary Points and Classification

Stationary points occur where $f'(x) = 0$. To classify them:

First derivative test: Check the sign of $f'(x)$ just before and after the stationary point.

• $f'$ changes from $+$ to $-$: local maximum
• $f'$ changes from $-$ to $+$: local minimum
• $f'$ does not change sign: point of inflection (horizontal)

Second derivative test: Find $f''(x)$ at the stationary point.

• $f''(a) < 0$: local maximum (concave down)
• $f''(a) > 0$: local minimum (concave up)
• $f''(a) = 0$: test is inconclusive — use the first derivative test instead

Optimisation

Optimisation problems ask you to find the maximum or minimum value of a quantity. The general approach:

1. Define variables and write the quantity to optimise as a function of one variable
2. Differentiate and set the derivative equal to zero
3. Solve for the critical value
4. Verify it is a maximum or minimum (using the second derivative or reasoning)
5. Answer the question in context

Often you need to use a constraint equation to eliminate one variable before differentiating.

Tip 1: Rewrite Before Differentiating

Before applying the power rule, rewrite expressions in index form: $\frac{1}{x^2} = x^{-2}$, $\sqrt[3]{x} = x^{1/3}$, $\frac{5}{\sqrt{x}} = 5x^{-1/2}$. This makes the power rule straightforward to apply.

Tip 2: Show the Rule You Are Using

In IB exams, examiners look for evidence that you know which rule to apply. Write "using the chain rule" or "by the product rule" — this earns method marks even if your arithmetic has a small error.

Tip 3: Use the Second Derivative for Classification

The second derivative test is usually quicker than the first derivative test. Compute $f''$ at the stationary point: if it is negative, you have a maximum; if positive, a minimum. Only resort to the sign-change test if $f'' = 0$.

Tip 4: Verify Optimisation Answers with Your GDC

After solving an optimisation problem algebraically, graph the function on your GDC and use the maximum/minimum feature to confirm your answer. This catches errors and takes only seconds.

Tip 5: Read Optimisation Questions Carefully

Make sure you answer what is asked. If the question asks for the minimum cost, don't stop after finding the value of $x$ — substitute back to find the actual cost.