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how do you differentiate this?? ln(sinxcosx) my working is d/dx[ln(sinxcosx)] =1/sinxcosx • (-cosxsinx) =-cosxsinx/cosxsinx =-1 but it looks pretty wrong..
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1/sinxcosxis right
bt after it we will apply u*v
d(sinxcosx)/dx=(cosx*dsinx/dx+sinx*dcosx/dx)
how about the ln?
ln is okkkk u did right
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wait no i don't see it... am i supposed to add your working to the one i did above? also why do you have the d and dxs all over the working? :o
ohhh okay so you did the product rule
\[\frac{ d \ln(sinxcosx)}{ dx}=\frac{ 1 }{ sinx*cosx }*\frac{ d(sinxcosx) }{ dx }\]
yeah product rule after diff.. ln
so i'll get cosx(cosx)+sinx(-sinx) =cos^2x-sin^2x?
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okay i got (cos^2x-sin^2x)/(sinxcosx) how do i simplify this?
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