# Finding derivative with log

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• Jul 14th 2013, 07:37 PM
JellyOnion
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.

the answer is tanx

any help appreciated, thx
• Jul 14th 2013, 08:05 PM
ibdutt
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)
• Jul 14th 2013, 08:22 PM
Soroban
Re: Finding derivative with log
Hello, JellyOnion!

Another way . . .

Quote:

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

$\displaystyle \frac{dy}{dx} \;\;=\;\;-\frac{1}{\cos x}(-\sin x) \;\;=\;\;\frac{\sin x}{\cos x} \;\;=\;\;\tan x$
• Jul 15th 2013, 09:02 AM
bobm2013
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.

For more info, please visit:-

Maths Discovery
• Jul 15th 2013, 09:32 AM
HallsofIvy
Re: Finding derivative with log
That's NOT the "product rule", that is the "chain rule".
• Jul 15th 2013, 10:11 AM
topsquark
Re: Finding derivative with log
Quote:

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.

For more info, please visit:-

I find this comment rather ironic since he's advertising a paid tutoring service.

-Dan
• Jul 16th 2013, 12:51 AM
JellyOnion
Re: Finding derivative with log
Lol, thx everyone. You guys helped heaps for my test today. Cheers