How do you differentiate ln((sin^2)x)?

2 Answers
Oct 30, 2016

Rewrite it first.

Explanation:

ln(sin^2 x) = ln((sinx)^2) = 2ln(sinx)

Now use d/dx(lnu) = 1/u du/dx, to get

d/dx(ln(sin^2 x)) = 2 d/dx(ln(sinx))

= 2(1/sinx)*d/dx(sinx)

= 2(1/sinx) (cosx)

= 2cotx

Oct 30, 2016

y=ln((sinx)^2)

Which means that:

e^y=(sinx)^2

Use implicit differentiation on the left hand side of the equation and the chain rule on the right hand side of the equation:

e^y*(dy)/(dx)=2sinx*cosx

Divide expressions on both sides of the equation by e^y:

(dy)/(dx)=(2sinx*cosx)/e^y

Don't forget that e^y is (sinx)^2:

(dy)/(dx)=(2sinxcosx)/(sinxsinx)

Simplify the fraction above:

(dy)/(dx)=2*cosx/sinx

cosx/sinx is cotx:

(dy)/(dx)=2cotx

Presto!!