How do you use the chain rule to differentiate #y=root5(x^2-3)/(-x-5)#?

1 Answer
Manikandan S.
Nov 21, 2017

Separate into components

Explanation:

#y = (root(5)(x^2-3))/(-x-5)#
Let #u = root(5)(x^2-3)# and #v =-1/(x+5)#

#y' = udv+vdu#
#y' = root(5)(x^2-3)\times1/(x+5)^2-1/(x+5)(1/5)(x^2-3)^(-4/5) 2x#
#y' = root(5)(x^2-3)\times1/(x+5)^2(1-(2x(x+5))/(x^2-3))#
#y' = root(5)(x^2-3)\times1/(x+5)^2(-(x^2+10x+3))/(x^2-3)#
#y' = (root(5)(x^2-3))^4 \times(1+22/(x+5)^2)#

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