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..

Question

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..

gorv · Answer

1/sinxcosxis right

gorv · Answer

bt after it we will apply u*v

gorv · Answer

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

anonymous · Answer

how about the ln?

gorv · Answer

ln is okkkk u did right

anonymous · Answer

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

anonymous · Answer

ohhh okay so you did the product rule

gorv · Answer

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

gorv · Answer

yeah product rule after diff.. ln

anonymous · Answer

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

anonymous · Answer

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