# Nth Derivative of a Function

Gold Member

## Homework Statement

Let ##f(x) = \frac{\sin x}{b + \cos(ax)}##. Show that the nth derivative ##f^{(n)}(0) = 0## if n is an even integer.

## Homework Equations

Leibniz's generalised product rule:

##(f \cdot g)^{(n)} = \sum_{k = 0}^{n} \binom{n}{k} f^{(k)}g^{(n-k)}##

## The Attempt at a Solution

I'm letting ##f(x) = \sin x## and ##g(x) = \frac{1}{b + \cos(ax)}## then applying Leibniz's rule. Clearly, the terms of the series k = 0, k = 2, ... (every even k) are all 0 when x = 0, since they all contain an even derivative of sin (which gives us sin again). But what do I do about the derivatives of g(x)? Is this the right approach?

Mathitalian
Odd functions of x have only odd powers of x in their Taylor-McLaurin series, so...

Gold Member
So you're saying I let ##\frac{1}{b + \cos(ax)} = a_0 + a_{1}x^2 + a_{2}x^4 + ... ##?