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

1 Answer
Jun 5, 2016

Answer:

#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)}#