Differentiation is a fundamental concept in calculus that allows us to analyse the rates at which things change. For students studying for their IGCSE exams, understanding how to apply differentiation to solve problems related to gradients, stationary points, and turning points is crucial. This set of notes will guide you through the process of applying differentiation, providing you with the tools and examples you need to grasp these concepts thoroughly.

Gradients of Curves

The gradient of a curve at any point is given by the derivative of its equation. For a curve described by $f(x) = ax^n$, the gradient is found by differentiating $f(x)$ to get $f'(x) = anx^{n-1}$.

• Example: For $f(x) = 3x^2$, differentiate to find the gradient: $f'(x) = 6x.$ This indicates the gradient of the curve at any point $x$ is $6x$.

Stationary Points

Stationary points occur where the gradient (or first derivative) of a curve is zero.

Finding Stationary Points

1. Differentiate the function $f(x)$ to get $f'(x)$.

2. Solve $f'(x) = 0$ for $x$ to find the stationary points.

Example

Given $f(x) = 3x^2 - 12x + 9$, we find the stationary point(s) as follows:

1. Differentiate $f(x)$ to get $f'(x) = 6x - 12$.

2. Solve for $x$ when $f'(x) = 0$: $6x - 12 = 0 \Rightarrow x = 2.$

3. Substitute $x = 2$ back into $f(x)$ to find $y$:

Hence, the stationary point is at $(2, -3)$.

Turning Points

Turning points are stationary points where the curve changes direction, identified by the second derivative $f''(x)$.

Identifying Turning Points

• If f''(x) > 0, the point is a minimum.
• If f''(x) < 0, the point is a maximum.

Example:

Find the stationary points and classify them for the function $g(x) = x^3 - 6x^2 + 9x + 1$.

Solution:

1. Differentiate the function: $g'(x) = 3x^2 - 12x + 9$.

2. Find stationary points: Solve $3x^2 - 12x + 9 = 0$.

Using the quadratic formula:

The values are $x = 1$ and $x = 3$.

3. Determine the nature of each stationary point by evaluating the second derivative $g''(x) = 6x - 12$.

• At $x = 1$, $g''(1) = 6(1) - 12 = -6$, indicating a maximum point.
• At $x = 3$, $g''(3) = 6(3) - 12 = 6$, indicating a minimum point.