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

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Answer 1 · 2015-11-06T11:09:30

Applying the chain rule, $\frac{d}{d x} ln ({sin}^{2} (x)) = 2 cot (x)$ $d/dxln(sin^2(x)) = 2cot(x)$

The chain rule states that $d/dxf(g(x)) = f'(g(x))*g'(x)$

Let $f(x) = ln(x)$ and $g(x) = sin^2(x)$

Then $f'(x) = 1/x$

To find $g'(x)$ we need to use the chain rule again with $g_1(x) = x^2$ and $g_2(x) = sin(x)$

Then $g_1'(x) = 2x$ and $g_2'(x) = cos(x)$
So, as $g(x) = g_1(g_2(x))$
$g'(x) = g_1'(g_2(x))*g_2'(x) = 2sin(x)*cos(x)$

Going back to the original problem, we have
$ln(sin^2(x)) = f(g(x))$

so, applying the chain rule, we get
$d/dxln(sin^2(x)) = f'(g(x))*g'(x) = 1/(sin^2(x))*2sin(x)cos(x)$

Finally, simplifying gives us the final result of

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

What is the derivative of f(x) = ln(sin^2x)f(x)=ln(sin2x)f(x) = ln(sin^2x)?