How do you use the chain rule to differentiate #y=(x^2+5x)^2+2(x^3-5x)^3#?

1 Answer
Henry W.
Oct 9, 2016

#(dy)/(dx)=2(2x+5)(x^2+5x)+6(3x^2-5)(x^3-5x)^2#

Explanation:

Chain rule: #(dy)/(dx)=(dy)/(du)*(du)/(dx)#

We do this twice to derive both #(x^2+5x)^2# and #2(x^3-5x)^3#

#d/(dx)(x^2+5x)^2#: Let #u=x^2+5x#, then #(du)/(dx)=2x+5#
#(dy)/(du)=2(x^2+5x)#
So #(dy)/(dx)=2(2x+5)(x^2+5x)#

#d/(dx)2(x^3-5x)^3#: Let #u=x^3-5x#, then #(du)/(dx)=3x^2-5#
#(dy)/(du)=6(x^3-5x)^2#
So #(dy)/(dx)=6(3x^2-5)(x^3-5x)^2#

Now adding both together,
#(dy)/(dx)=2(2x+5)(x^2+5x)+6(3x^2-5)(x^3-5x)^2#

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