- Gradient of a line passing through two points
- Limits
**Differentiation from first principles**- Differentiating expressions of the form \(kx^n\) with respect to \(x\)
- The gradient at a point on a curve
- Tangents and normals

# Part 3: Differentiation from first principles

### Finding the gradient at a point on the curve \(y=x^2\)

Given a curve \(y=\text{f}(x)\), for certain functions \(\text{f}\), we can find the **derivative** or **gradient function** of the curve. The derivative is a function that allows us to find the gradient at any point on the original curve.

Work through this interactive applet showing how to find the gradient function for the curve \(y=x^2\).

#### Vocabulary and notation

The **derivative** or gradient function is a function that allows us to find the gradient at any point on the original curve.

The *process* of finding the derivative or gradient function is known as **differentiation**. When looking for the gradient in the \(x\)-\(y\) plane, we differentiate *with respect to* \(x\) to find the derivative *with respect to* \(x\).

The derivative of \(y\) with respect to \(x\) is written \(\dfrac{\text{d}y}{\text{d}x}\). We say this as “\(\text{d}y\) by \(\text{d}x\)”. In the above example, we saw that differentiating \(x^2\) with respect to \(x\) gave us \(2x\).

We also have some alternative notation we can use when working with functions: the derivative of \(\text{f}(x)\) with respect to \(x\) is denoted \(\text{f}'(x)\). We say this as “\(\text{f}\) prime of \(x\)” or “\(\text{f}\) dash of \(x\)”.

**Questions**

- If \(y=x^3\), find \(\dfrac{\text{d}y}{\text{d}x}\).
- If \(y=5x^2\), find \(\dfrac{\text{d}y}{\text{d}x}\).
- If \(y=x^3+5x^2\), find \(\dfrac{\text{d}y}{\text{d}x}\).
- If \(\text{f}(x)=x^4\), find \(\text{f}'(x)\).

**Answers**

- If \(y=x^3\), \(\dfrac{\text{d}y}{\text{d}x}=3x^2\).
- If \(y=5x^2\), \(\dfrac{\text{d}y}{\text{d}x}=10x\).
- If \(y=x^3+5x^2\), \(\dfrac{\text{d}y}{\text{d}x}=3x^2+10x\).
- If \(\text{f}(x)=x^4\), \(\text{f}'(x)=4x^3\).