Question

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

Answer 1

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

Answer 2

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?

Answer 3

I mean not only NOT necessary...

Answer 4

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

Answer 5

y=ln(( sinx)^2)

y' = csc^2 x cosx

Answer 6

no, hold... I am wrong

Answer 7

sin^2x = (sinx)^2

d/dx = 2sinx cosx


then
ln((sinx)^2) = csc^2x * 2sinx cosx = cot x

right?

Answer 8

2 cotx (sorry)

Answer 9

You got 2cot(x) either way.

Answer 10

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