# Math Help - Quotient rule with Sin & Cos

1. ## Quotient rule with Sin & Cos

f(x)/g(x) = f'(x) g(x) - f(x) g'(x) / g(x)^2
How do I find the derivative of something like cos(x) / sin(1/x) ?

= ( - sin x) sin(1/x) - cos (x) cos ( 1/x) / ( sin( 1/x )^ 2

I know this is not completely correct. What is the derivative of sin ( 1/x )? Do I leave the value as 1/x ?

2. Originally Posted by bt12345
f(x)/g(x) = f'(x) g(x) - f(x) g'(x) / g(x)^2
How do I find the derivative of something like cos(x) / sin(1/x) ?

= ( - sin x) sin(1/x) - cos (x) cos ( 1/x) / ( sin( 1/x )^ 2

I know this is not completely correct. What is the derivative of sin ( 1/x )? Do I leave the value as 1/x ?
$g(x)=sin(\frac{1}{x})$ is a a composite function, so you have to use the chain rule. Are you familiar with this?

$g'(x)=[\frac{d}{dx}(\frac{1}{x})]cos(\frac{1}{x})=-\frac{1}{x^2}cos(\frac{1}{x})$

In general, if you have a composite function:

$F(x)=f(g(x))$

The derivative of that function is:

$F'(x)=f'(g(x))g'(x)$