What is the derivative of #(ln(sin(2x)))^2#?

Answer 1

#d/dx((ln(sin(2x)))^2)=4cot2xln(sin(2x))#

If we let #y=(ln(sin(2x)))^2#, then we can find the derivative of #y# using the chain rule. But first, we will have to define the composition of the functions we need to differentiate.

Let #y=u^2#

and #u=lnv#

and #v=sinw#

and #w=2x#

Then,

#dy/dx=dy/(du)xx(du)/(dv)xx(dv)/(dw)xx(dw)/dx#

#dy/(du)=2u=2lnv=2ln(sinw)=2ln(sin(2x))#

#(dv)/(du)=1/v=1/sinw=1/sin(2x)#

#(dv)/(dw)=cosw=cos(2x)#

#(dw)/dx=2#

#dy/dx=2ln(sin(2x))xx1/sin(2x)xxcos(2x)xx2#

#=4cot(2x)ln(sin(2x))#

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Answer 2

Evelyn Agard

The derivative of ( (\ln(\sin(2x)))^2 ) is ( 4 \ln(\sin(2x)) \cdot \frac{\cos(2x)}{\sin(2x)} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.