This section is dedicated to exploring the practical applications of differentiation in the field of mathematics. We will look into how differentiation is used to calculate gradients of curves, derive equations for tangents and normals, and understand when a function is increasing or decreasing. We will also solve problems involving rates of change, including connected rates, such as changes in radius and area of a circle.

**Finding the Gradient Using Differentiation**

**Gradient of a Curve**

The derivative $f'(x)$ of a function $y = f(x)$ represents the gradient of the curve at any point $x$ on the graph.

**Tangents and Normals**

- The gradient of the tangent at a point on a curve is obtained by differentiating the function and substituting the $x$-coordinate of that point.
- The gradient of the normal at a point is the negative reciprocal of the tangent's gradient, as normals are perpendicular to tangents.

**Equations of Tangents and Normals**

With knowledge of the gradient of a tangent or normal at a point on a curve, one can derive the equation of the line utilizing the formula $y = mx + c$. By inserting the coordinates of the point into this formula, the value of $c$ can be ascertained, thereby finalizing the equation.

**Examples**

**Example 1: Gradient of Tangent and Normal**

Consider the curve $y = 3x^2$. Determine the gradient of the tangent and the normal at the point where $x = 2$.

**Solution:**

**1. Gradient of the Tangent**:

- Differentiate the function: $\frac{dy}{dx} = 6x$.
- At $x = 2$, the gradient is $6 \times 2 = 12$.
- Therefore, the gradient of the tangent at $x = 2$ is 12.

**2. Gradient of the Normal**:

The gradient of the normal is $-\frac{1}{12}$ (negative reciprocal of tangent's gradient).

**Example 2: Gradient and Equation of Tangent**

Find the gradient and equation of the tangent to the curve $y = x^3 - 2x$ at the point where $x = 1$.

**Solution**

**1. Gradient of the Tangent**:

- Differentiate the function: $\frac{dy}{dx} = 3x^2 - 2$.
- At $x = 1$, the gradient is $3(1)^2 - 2 = 1$.

**2. Equation of the Tangent**:

Using the point-slope form, the equation is $y = x - 2$.