Mastering Mathematics Smartly
by Wee Wen Shih

A unique self-help website that provides comprehensive coverage of mathematics at A-level & beyond, written in a student-friendly style.

Proving nth derivative

Mathematical induction can be applied to prove results involving the n-th derivative of a function. The steps that we know of in H2 mathematics are still the same.

Let's use mathematical induction to prove the result: if y = e^x sin x then \frac{d^ny}{dx^n} = 2^{\frac{1}{2}n} e^x sin\left(x + \frac{1}{4}n\pi \right).

Let the statement be \frac{d^ny}{dx^n} = 2^{\frac{1}{2}n} e^x sin\left(x + \frac{1}{4}n\pi \right) for n \geq 1.

When n = 1,
\\\text{LHS} \\= \frac{d}{dx}\left(e^x sin x\right) \\= e^x cos x + e^x sin x \\=e^x\left[\sqrt{2} sin \left(x + \frac{\pi}{4} \right) \right] \\=\text{RHS, when }n = 1

Therefore the statement is true n = 1.

Assume that the statement is true for n = k, i.e.
\frac{d^ky}{dx^k} = 2^{\frac{1}{2}k} e^x sin\left(x + \frac{1}{4}k\pi \right).

We want to show that the statement is true for n = k+1, i.e.
\frac{d^{k+1}y}{dx^{k+1}} = 2^{\frac{1}{2}(k+1)} e^x sin\left(x + \frac{1}{4}(k+1)\pi \right)

To obtain the (k + 1)-th derivative, we differentiate \frac{d^ky}{dx^k} one more time.

\text{LHS} = \frac{d}{dx}\left(\frac{d^ky}{dx^k}\right)
= 2^{\frac{1}{2}k}\left[e^x cos\left(x + \frac{1}{4}k\pi\right) + e^x sin\left(x + \frac{1}{4}k\pi\right) \right]

= 2^{\frac{1}{2}k}\left[e^x \sqrt{2} sin\left(x + \frac{1}{4}k\pi + \frac{1}{4}\pi\right)\right], which simpliflies to the desired result.

Therefore the statement is true for n = k+1 if it is true for n = k.

Hence the statement is true by mathematical induction.

Proving inequalities

Induction can be applied to prove results involving inequalities too.

Let's look at a question asked by someone on Yahoo! Answers:

To prove that the result holds when n = 1 is straightforward. I will do the portion which is more difficult due to the need for manipulations.