How do you find the derivative of # (y - 1)^5 / (y^2 + 5 y)^6# using the chain rule?

Question

1921 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-20T21:51:23

The main tool for differentiating this is the Quotient Rule ; the chain rule is secondary. Please see the explanation.

Let #f(y) =g(y)/(h(y)) = (y - 1)^5/(y^2 + 5y)^6#

#f'(y) = {g'(y)h(y) - g(y)h'(y)}/(h(y))^2#

let #g(y) = (y - 1)^5#

Here is where the chain rule comes in:

Let #u(y) = y - 1#

#g'(u(y)) = g'(u)(u'(y))#

#u'(y) = 1#

#g'(u(y)) = g'(u) = 5(y - 1)^4#

#h = (y^2 + 5y)^6#

Again the chain rule is used for h'(y):

#h'(u(y)) = h'(u)(u'(y))#

let #u(y) = y^2 + 5y, u'(y) = 2y + 5#

#h'(u(y)) = h'(u)(u'(y))#

#h'(u(y)) = 6(y^2 + 5y)^5(2y + 5)#

You now have all of the parts to substitute back into the quotient rule:

#f'(y) = {( 5(y - 1)^4)(y^2 + 5y)^6 - (y - 1)^5(6(y^2 + 5y)^5(2y + 5))}/(y^2 + 5y)^12#

#f'(y) = {( 5(y - 1)^4)(y^2 + 5y) - (y - 1)^5(6(2y + 5))}/(y^2 + 5y)^7#

#f'(y) = {( 5(y - 1)^4)(y^2 + 5y) - (y - 1)^5(12y + 30)}/(y^2 + 5y)^7#