How do you differentiate f(x)=cot(4x-x^2) using the chain rule?

1 Answer
Aniket Mahaseth
Jun 5, 2016

frac{d}{dx}(cot(4x-x^2))=-frac{-2x+4}{sin ^2(4x-x^2)}

Explanation:

frac{d}{dx}(cot(4x-x^2))
Applying chain rule,frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}
Let, 4x-x^2 =u
=frac{d}{du}(cot (u))frac{d}{dx}(4x-x^2)

We know,
frac{d}{du}(cot (u))=-frac{1}{sin ^2(u)}
and,
frac{d}{dx}(4x-x^2)=4-2x

So,
=(-frac{1}{sin ^2(u)})(4-2x)

Substituting back, 4x-x^2 =u

=(-frac{1}{sin ^2(4x-x^2)})(4-2x)

Simplifying it,
-frac{-2x+4}{sin ^2(4x-x^2)}

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