# Math Help - Finding derivative with log

1. ## Finding derivative with log

Can't seem to get this question.

find dy/dx of y=-log(base,e)(cosx)

i get the expression to (-cosx/x)+log(base,e)+sinx. But I don't know how to take it any further.

any help appreciated, thx

2. ## Re: Finding derivative with log

You are probably going wrong
we have y = - log cosx = log ( cosx)^-1= log sec x
dy/dx = (1/sec x )( sec x tan x ) = tan x
remember differential of log ( f(x)) = (1/f(x) ) f'(x)

3. ## Re: Finding derivative with log

Hello, JellyOnion!

Another way . . .

$\text{Differentiate: }\:y \:=\:-\ln(\cos x)$

$\frac{dy}{dx} \;\;=\;\;-\frac{1}{\cos x}(-\sin x) \;\;=\;\;\frac{\sin x}{\cos x} \;\;=\;\;\tan x$

4. ## Re: Finding derivative with log

What you need is the product rule. So d/dx (f(g(x))) is f'(g) x g'(x). log'(f) is 1/f, d/dx(cos) is -sin,
so you have -sin/-cos , which is tan.

Maths Discovery

5. ## Re: Finding derivative with log

That's NOT the "product rule", that is the "chain rule".

6. ## Re: Finding derivative with log

Originally Posted by bobm2013
What you need is the product rule. So d/dx (f(g(x))) is f'(g) x g'(x). log'(f) is 1/f, d/dx(cos) is -sin,
so you have -sin/-cos , which is tan.