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

Question

2834 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-25T08:44:13

#d/dxln(sin^2(x)) = 2cot(x)#

We will use the following:
The chain rule
#d/dxf(g(x)) = f'(g(x)g'(x)#

#d/dxln(x) = 1/x#

#d/dx x^n = nx^(n-1)#

#d/dx sin(x) = cos(x)#

Now, as the function given is a logarithm of a power of the sine function, we will apply the chain rule twice:

#d/dxln(sin^2(x)) = 1/(sin^2(x))*(d/dxsin^2(x))#

#=> d/dxln(sin^2(x)) = 1/(sin^2(x))* 2sin(x)*(d/dx sin(x))#

#=> d/dxln(sin^2(x)) = (2sin(x))/(sin^2(x))cos(x) = (2cos(x))/sin(x)=2cot(x)#