In this section, our focus will be on differentiating powers of x with rational exponents, as well as employing the chain rule for composite functions. These techniques are essential for understanding how functions change and evolve.

**Differentiating Powers of x with Rational Exponents**

When working with x raised to a rational exponent, the differentiation process follows a standard rule. For $f(x) = x^n$, where $n$ is a rational number, the derivative is $f'(x) = nx^{n-1}$.

**Example**

Differentiate $y = x^{3/2}$.

**Solution:**

**Differentiation of Sums, Differences, and Constant Multiples of Functions**

These rules enable simplification before applying differentiation:

1. **Sum:** $\frac{d}{dx}(u(x) + v(x)) = u'(x) + v'(x)$

2. **Difference:** $\frac{d}{dx}(u(x) - v(x)) = u'(x) - v'(x)$

3. **Constant Multiple:** $\frac{d}{dx}(c \cdot u(x)) = c \cdot u'(x)$

**Example**

Differentiate $y = 3x^2 + 4x$.

**Solution:**

**Chain Rule**

This rule is used for differentiating composite functions and is formulated as:

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

Or,

$f(g(x)))' = f'(g(x)) \times g'(x)$

**Example**

Differentiate $y = (x + x^3)^5$.

**Solution:**

Let $u = x + x^3$, then find $\frac{du}{dx}$

$u = x + x^3$$\frac{du}{dx} = 1 + 3x^2$Now ( y = u^5 )

$\frac{dy}{du} = 5u^4$Multiply them together:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = (1 + 3x^2) \times 5(x + x^3)^4$Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.