[InternetGangster]

Problem solved.

[Augmented hippo]

If y=ln(f(x)) then dy/dx=f'(x)/f(x)
In this case f(x) is sinx and f'(x) is cosx therefore dy/dx=cos(x)/sin(x) or cot(x).

[notnek]

The derivative of ln(f(x)) is the derivative of f(x) divided by f(x):



So what is dy/dx where f(x)=sin(x)?

[kingkongjaffa]

chain rule buddy

differentiate what's inside the outer function so differentiate ln(sinx) to give 1/sinx then multiply by the derivative of the inner function (sinx))

what's the derivative of sinx?

[InternetGangster]

(Original post by kingkongjaffa)
chain rule buddy

differentiate what's inside the outer function so differentiate ln(sinx) to give 1/sinx then multiply by the derivative of the inner function (sinx))

what's the derivative of sinx?

Oh yeah, thanks. I was trying to use the product rule to differentiate instead.

[InternetGangster]

(Original post by Augmented hippo)
If y=ln(f(x)) then dy/dx=f'(x)/f(x)
In this case f(x) is sinx and f'(x) is cosx therefore dy/dx=cos(x)/sin(x) or cot(x).

thank you

[InternetGangster]

(Original post by notnek)
The derivative of ln(f(x)) is the derivative of f(x) divided by f(x):



So what is dy/dx where f(x)=sin(x)?

Thanks