# What is the nth derivative of sin (ax +b)cos(ax+b)?

##### 2 Answers
Feb 8, 2015

Here we shall make use of the product rule, which states that $\frac{d}{\mathrm{dx}} \left(f \left(x\right) g \left(x\right)\right) = \frac{\mathrm{df}}{\mathrm{dx}} g \left(x\right) + f \left(x\right) \frac{\mathrm{dg}}{\mathrm{dx}}$
$= f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

We will begin by finding the first derivative.

*Note: The below assumes the student is comfortable with the chain rule, the trigonometric identities, and the trigonometric derivatives. *

Here, $f \left(x\right) = \sin \left(a x + b\right)$ and $g \left(x\right) = \cos \left(a x + b\right)$. Via use of the chain rule and the definitions for trigonometric derivatives, this yields $f ' \left(x\right) = a \cos \left(a x + b\right)$ and $g ' \left(x\right) = - a \sin \left(a x + b\right)$. Thus, our derivative is

$f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right) = a \cos \left(a x + b\right) \cos \left(a x + b\right) + \left(- 1\right) a \sin \left(a x + b\right) \sin \left(a x + b\right)$
$= a \left({\cos}^{2} \left(a x + b\right) - {\sin}^{2} \left(a x + b\right)\right)$
We know from the double angle identities that $\cos \left(2 u\right) = {\left(\cos u\right)}^{2} - {\left(\sin u\right)}^{2} = {\cos}^{2} \left(u\right) - {\sin}^{2} \left(u\right)$. Thus, for $u = a x + b$...

f'(x)g(x) + f(x)g'(x) = a (cos^2 u - sin^2 u) = a cos (2u) = a cos (2ax+2b)

This only yields the first derivative; however, from here, determining the later derivatives is a matter of recalling the angle-sum equations, specifically $\cos \left(a + b\right) = \cos \left(a\right) \cos \left(b\right) - \sin \left(a\right) \sin \left(b\right)$. For $h \left(x\right) = \cos \left(x\right) , h ' \left(x\right) = - \sin \left(x\right) , h \text{(x) = -cos(x), h"'(x) = sin(x), h} \left(x\right) = \cos \left(x\right)$, and so forth.

Using our sum identity, we determine that $\cos \left(\frac{\pi}{2} + x\right) = - \sin \left(x\right) , \cos \left(\pi + x\right) = - \cos \left(x\right) , \cos \left(\frac{3}{2} \pi + x\right) = \sin \left(x\right) , \mathmr{and} \cos \left(2 \pi + x\right) = \cos \left(x\right)$ Thus, the ${n}^{t h}$ derivative of $\cos \left(x\right) = {h}^{\left(n\right)} x = \cos \left(\frac{n \pi}{2} + x\right)$ In our case, however, we have $h \left(x\right) = a \cos \left(2 a x + 2 b\right)$, and thus - via the Chain Rule - our ${n}^{t h}$ derivative would actually be ${h}^{\left(n\right)} x = {2}^{n} {a}^{n + 1} \cos \left(\frac{n \pi}{2} + x\right)$

Note also, however, that the ${n}^{t h}$ derivative of $h \left(x\right)$ is actually the ${\left(n + 1\right)}^{t h}$ derivative of our initial function. To find the ${n}^{t h}$ derivative of the initial function, we need to find the ${\left(n - 1\right)}^{t h}$ derivative for $h \left(x\right)$ Thus, for our initial function $f \left(x\right) g \left(x\right) = j \left(x\right)$...
${j}^{\left(n\right)} \left(x\right) = {2}^{n - 1} {a}^{n} \cos \left(\frac{\left(n - 1\right) \pi}{2} + x\right)$

Aug 11, 2015

#### Answer:

${f}^{\left(n\right)} \left(x\right) = {\left(2 a\right)}^{n} / 2 \sin \left[2 \left(a x + b\right) + \frac{n \pi}{2}\right] , \text{ } n \in {\mathbb{Z}}_{0}^{+}$

#### Explanation:

$\sin 2 A = 2 \sin A \cos A$
$\cos A = \sin \left(A + \frac{\pi}{2}\right)$
$- \sin A = \sin \left(A + \pi\right)$

$f \left(x\right) = \sin \left(a x + b\right) \cos \left(a x + b\right) = \frac{1}{2} \sin 2 \left(a x + b\right)$
$f ' \left(x\right) = \frac{\mathrm{df}}{\mathrm{dx}} = a \cos 2 \left(a x + b\right) = a \sin \left[2 \left(a x + b\right) + \frac{\pi}{2}\right]$
$f ' ' \left(x\right) = - 2 {a}^{2} \sin 2 \left(a x + b\right) = 2 {a}^{2} \sin \left[2 \left(a x + b\right) + \pi\right]$
${f}^{\left(3\right)} \left(x\right) = - 4 {a}^{3} \cos 2 \left(a x + b\right) = 4 {a}^{3} \sin \left[2 \left(a x + b\right) + \frac{3 \pi}{2}\right]$

${f}^{\left(n\right)} \left(x\right) = {\left(2 a\right)}^{n} / 2 \sin \left[2 \left(a x + b\right) + \frac{n \pi}{2}\right] , \text{ } n \in {\mathbb{Z}}_{0}^{+}$