How do you find the derivative of #y=ln(sin2x)#?

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Answer 1 · 2018-03-31T06:19:21

The derivative is #2cot(2x)#.

We'll need to use the derivative chain rule:

#color(white)=d/dx(f(g(x)))=f'(g(x))*g'(x)#

where our #f(x)# is #ln(x)# and our #g(x)# is #sin2x#.

This means that #f'(x)# would be #1/x# and #g'(x)# would need the chain rule again:

#color(white)=d/dx(sin(2x))#

#=sin'(2x)*d/dx(2x)#

#=cos(2x)*2#

#=2cos(2x)#

So our #f'(x)# is #1/x# and #g'(x)# is #2cos(2x)#.

Plugging in the functions:

#color(white)=d/dx(ln(sin2x))#

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

#=(2cos(2x))/sin(2x)#

#=2*cos(2x)/sin(2x)#

#=2*cot(2x)#

#=2cot(2x)#

That's the derivative. Hope this helped!