Basic Derivative?

RJC

Alright, here is the equation:

ln [(sin(x)^3)^3]

And here is my answer with some steps, please tell me if i'm wrong..

9 ln(sin'x)

9 (cos'x)/(sin'x)

Derivative= 9tan(x)^-1

is that correct?

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mathmike

I Get 9*x^2*cot(x^3)
It Looks Like U Forgot The X^2 Term When You Got The Der Of X^3

Jameson

Usually we say that $$\frac{1}{\tan(x)}=\cot(x)$$. Looks good to me.

RJC

soo.. which answer is correct? the whole sin'x is cubed, not just the x value within the sine..

Jameson

I see nothing wrong with your original answer, except for a better notation.

RJC

so 9cot(x) would be a better answer?

Jameson

Yes, it's more concise and very standard.

"Basic Derivative?"

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