Question

find dy/dx for the following function

y = 9cosx /(1+sinx)

u=9cosx v= 1+six
u' = -9sinx v' = cosx

u'v - v'u / (v)^2

-9sinx (1+sinx)- cosx(9cosx) / (1+sinx)^2 ?

not sure if i'm doing this right... answer is -(9/1+sinx) but I'm not getting that.

Answer 1

You forgot your trig identifty for \[-9\sin ^{2}x-9\cos^{2}x=-9(\sin ^{2}x +\cos ^{2}x)=-9\]

Then your top simplifies to:

\[-9sinx-9=-9(sinx+1)\]

So, one of the sinx+1's in the denominator will cancel with the top one, leaving your answer as required.

Answer 2

Thank you!

Answer 3

You are welcome.