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

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Answer 1 · 2016-05-06T07:06:44

#d y=-(2x-4)*sin(x^2-4x)*d x#

#f(x)=y=cos(x^2-4x)#

#u=x^2-4x#

y=cos u##

#(d u)/(d x)=2x-4#

#(d y)/(d u)=-sin u=-sin(x^2-4x)#

#(d y)/(d x)=(d u)/(d x)*(d y)/(d u)#

#(d y)/(d x)=-(2x-4)*sin(x^2-4x)#

#d y=-(2x-4)*sin(x^2-4x)*d x#