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OpenStudy (anonymous):

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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OpenStudy (gorv):

1/sinxcosxis right

OpenStudy (gorv):

bt after it we will apply u*v

OpenStudy (gorv):

d(sinxcosx)/dx=(cosx*dsinx/dx+sinx*dcosx/dx)

OpenStudy (anonymous):

how about the ln?

OpenStudy (gorv):

ln is okkkk u did right

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OpenStudy (anonymous):

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

OpenStudy (anonymous):

ohhh okay so you did the product rule

OpenStudy (gorv):

\[\frac{ d \ln(sinxcosx)}{ dx}=\frac{ 1 }{ sinx*cosx }*\frac{ d(sinxcosx) }{ dx }\]

OpenStudy (gorv):

yeah product rule after diff.. ln

OpenStudy (anonymous):

so i'll get cosx(cosx)+sinx(-sinx) =cos^2x-sin^2x?

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OpenStudy (anonymous):

okay i got (cos^2x-sin^2x)/(sinxcosx) how do i simplify this?

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