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

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Answer 1 · 2016-10-30T19:33:03

Rewrite it first.

#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#

Answer 2 · 2016-10-30T19:42:06

#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!!