how do i find the derivative of f(x) = log base 10 (2x^3 - 4x) + ln(sin^2x)?

Question

how do i find  the derivative of f(x) = log base 10 (2x^3 - 4x) + ln(sin^2x)?

anonymous · Answer

chain rule

anonymous · Answer

$$\frac{d}{dx}\log_b(f(x))=\frac{f'(x)}{\ln(b) f(x)}$$

anonymous · Answer

and $$\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}$$

anonymous · Answer

what will the answer be?...

lalaly · Answer

Try to find it ..

anonymous · Answer

To start, split it into the two separate parts: $ \log_{10}(2x^3-4x) $ and $\ln(\sin^2x)$, then derive those parts each using the two formulas that satellite gave you.

anonymous · Answer

i already did the problem am checkin to see if i got the right answer....
i got  2 cot x + (4sec^2/tan x) - 4x/(x^2 + 1)

anonymous · Answer

That's not at all what I got. Can you show your work?

anonymous · Answer

The $2\cot x$ part is right, but I don't know where you got the rest from.

anonymous · Answer

my teacher marked it correct..

anonymous · Answer

can you show your workout? thank

lalaly · Answer

$$\frac{d}{dx}(\log_{10}(2x^3-4x)+\ln(\sin^2x))$$$$=\frac{6x^2-4}{\ln(10)(2x^3-4x)}+\frac{2cosxsinx}{\sin^2x}$$$$=\frac{2(3x^2-2)}{2\ln(10)(x^3-2)}+2cotx$$

anonymous · Answer

thanks this looks close to what  i got..