# Further Differentiation and Applications

> CIE A-Level Further Mathematics · 9231 Further Pure 2
> Source: https://www.owlsprep.com/study/cie-9231-u2-further-differentiation-and-applications/

This sub-topic extends basic differentiation to advanced techniques including parametric, implicit, and higher-order differentiation, with applications to series expansions and stationary point classification. These methods are core to further pure mathematics and appear frequently in both papers.

**Prerequisites:** [Basic differentiation, product/chain rule](https://www.owlsprep.com/study/cie-9231-u1-basic-differentiation/); Parametric equations; Binomial coefficients and factorials

## Learning objectives

- Differentiate parametric, implicit, and inverse functions
- Find nth derivatives of products using Leibniz's theorem
- Construct Maclaurin and Taylor series expansions of functions
- Classify stationary points using higher-order derivatives

## Advanced Differentiation Techniques

Beyond basic differentiation of explicit functions, we extend techniques to relations defined implicitly and curves given by parametric equations, which are common in CIE exam questions.

**Parametric Differentiation** — The derivative of $y$ with respect to $x$ is found by dividing the derivative of $y$ with respect to $t$ by the derivative of $x$ with respect to $t$:

*Notation:* $x = x(t), y = y(t)$

*Example:* For $x = t^2, y = \sin t$, $\frac{dy}{dx} = \frac{\cos t}{2t}$

**Worked example:** Find $\frac{dy}{dx}$ for the ellipse defined parametrically by $x = 2\cos\theta$, $y = 3\sin\theta$, at the point where $\theta = \frac{\pi}{4}$.

1. First differentiate $x$ and $y$ individually with respect to $\theta$:
2. $$\frac{dx}{d\theta} = -2\sin\theta, \quad \frac{dy}{d\theta} = 3\cos\theta$$
3. Apply the parametric differentiation formula $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$:
4. $$\frac{dy}{dx} = \frac{3\cos\theta}{-2\sin\theta} = -\frac{3}{2}\cot\theta$$
5. Substitute $\theta = \frac{\pi}{4}$, where $\cot(\frac{\pi}{4}) = 1$:
6. $$\frac{dy}{dx}\bigg|_{\theta=\frac{\pi}{4}} = -\frac{3}{2}}]}}},{$$

Leibniz's theorem generalizes the product rule to nth derivatives, and is commonly used to find coefficients for Maclaurin series.

**Leibniz's Theorem** — The nth derivative of the product is given by the binomial-like sum:

*Notation:* $y = u(x)v(x)$

$$y^{(n)} = \sum_{k=0}^n \binom{n}{k} u^{(k)} v^{(n-k)}}{}, {}}, {}}, {{$$

**Worked example:** Find the third derivative of $y = x^2 \sin x$ using Leibniz's theorem.

1. Split the product into $u = x^2$ (all derivatives higher than 2 are zero, simplifying the sum) and $v = \sin x$. The derivatives of $u$ are: $u = x^2$, $u' = 2x$, $u'' = 2$, $u^{(k)} = 0$ for $k \geq 3$. The nth derivative of $v = \sin x$ is $v^{(n)} = \sin\left(x + \frac{n\pi}{2}\right)$.
2. Substitute into Leibniz's theorem for $n=3$:
3. $$y^{(3)} = \binom{3}{0}u v^{(3)} + \binom{3}{1}u' v^{(2)} + \binom{3}{2}u'' v^{(1)} + 0$$
4. Substitute values for each term:
5. $$y^{(3)} = x^2(-\cos x) + 3(2x)(-\sin x) + 3(2)(\cos x)$$
6. Simplify to get the final result:
7. $$y^{(3)} = (6 - x^2)\cos x - 6x \sin x}]}], {}}, {}}, {}},$$

*Calculator:* forbidden

*Calculator:* forbidden

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