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How do you find the derivative of #(ln(sin(2x)))^2#?

Answer 1

Evelyn Agard

#f'(x)=4ln(sin(2x))*cot(2x)#

Chain rule!

#f(x)=(ln(sin(2x)))^2#

let #u=ln(sin(2x))# now #f(x)=u^2#

so #f'(x)=2u*(du)/dx = 2ln(sin(2x))*(du)/dx#

now to find #(du)/dx#

let #w=sin(2x)# now #u=ln(w)#

so #(du)/dx=1/w*(dw)/(dx) = 1/sin(2x)*(dw)/(dx)#

now to find #(dw)/(dx)#

let #s=2x# now #w=sin(s)#

so #(dw)/(dx)=cos(s)*(ds)/(dx) = cos(s)*(ds)/(dx)#

We know that #(ds)/dx = 2# so...

#(dw)/(dx)=cos(2x)*2#

#(du)/dx= 1/sin(2x)*cos(2x)*2#

#f'(x)=2ln(sin(2x))*1/sin(2x)*cos(2x)*2#

simplify

#f'(x)=4ln(sin(2x))*cos(2x)/sin(2x)#

#f'(x)=4ln(sin(2x))*cot(2x)#

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Answer 2

Paisley Johnson

To find the derivative of ( (\ln(\sin(2x)))^2 ), you can use the chain rule and the power rule for differentiation. First, differentiate the outer function ( u^2 ), where ( u = \ln(\sin(2x)) ), then multiply by the derivative of the inner function ( \ln(\sin(2x)) ).

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Answer from HIX Tutor

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