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

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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:

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

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

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

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