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

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- Jul 14th 2013, 07:37 PMJellyOnionFinding 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 PMibduttRe: 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 PMSorobanRe: 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 AMbobm2013Re: 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 AMHallsofIvyRe: Finding derivative with log
That's NOT the "product rule", that is the "chain rule".

- Jul 15th 2013, 10:11 AMtopsquarkRe: Finding derivative with log
- Jul 16th 2013, 12:51 AMJellyOnionRe: Finding derivative with log
Lol, thx everyone. You guys helped heaps for my test today. Cheers