for the derivative of
y=ln(sin^2x)
do I apply chain twice?

Question

for the derivative of
y=ln(sin^2x)
do I apply chain twice?

anonymous · Answer

does it help to know that 
$$\ln(\sin^2(x))=2\ln(\sin(x))$$?

anonymous · Answer

ohh... yeah, taking the exponent out is nice:)
But just asking, would it be wrong (not only necessary, but also wrong) to apply chain twice?

anonymous · Answer

I mean not only NOT necessary...

anonymous · Answer

try it and see what happens
you will get a longer expression, then some trig identity will probably give you the shorter version

anonymous · Answer

y=ln(( sinx)^2)

y' = csc^2 x  cosx

anonymous · Answer

no, hold... I am wrong

anonymous · Answer

sin^2x  = (sinx)^2 

d/dx = 2sinx cosx


then
ln((sinx)^2) = csc^2x * 2sinx cosx = cot x

right?

anonymous · Answer

2 cotx  (sorry)

aum · Answer

You got 2cot(x) either way.

anonymous · Answer

cool... just wierd to apply the chain twice... tnx for the help everyone!