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

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Answer 1 · 2015-11-06T11:09:30

Applying the chain rule, #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)#