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Let f(x) = cotx

Then we know that: cot x = cosx/sinx

==> f(x) = cosx/sinx

Let us differetiate using the chain rule:

f(x) = u/v   such that:

u= cosx  =-=> u' = -sin

v= sinx ==> v' = cosx

==> f'(x) = (u'v- uv')/v^2

                = (-sinx*sinx - cosx*cosx)/...

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Let f(x) = cotx

Then we know that: cot x = cosx/sinx

==> f(x) = cosx/sinx

Let us differetiate using the chain rule:

f(x) = u/v   such that:

u= cosx  =-=> u' = -sin

v= sinx ==> v' = cosx

==> f'(x) = (u'v- uv')/v^2

                = (-sinx*sinx - cosx*cosx)/ sin^2 x

                  = (-sin^2x - cos^2x)/sin^2 x

Let us fctor -1 from the numerator:

==> f'(x) = -1(sin^2x + cos^2 x)/sin^2 x

We know that:

sin^2 x + cos^2 x = 1

==> f'(x) = -1/sin^2 x

But csc x = 1/sinx

==> f'(x) = - csc^2 x ...... q.e.d 

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