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

Oct 20, 2016

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

#### Explanation:

Let $f \left(y\right) = g \frac{y}{h \left(y\right)} = {\left(y - 1\right)}^{5} / {\left({y}^{2} + 5 y\right)}^{6}$

The Quotient Rule is:

$f ' \left(y\right) = \frac{g ' \left(y\right) h \left(y\right) - g \left(y\right) h ' \left(y\right)}{h \left(y\right)} ^ 2$

let $g \left(y\right) = {\left(y - 1\right)}^{5}$

Here is where the chain rule comes in:

Let $u \left(y\right) = y - 1$

$g ' \left(u \left(y\right)\right) = g ' \left(u\right) \left(u ' \left(y\right)\right)$

$u ' \left(y\right) = 1$

$g ' \left(u \left(y\right)\right) = g ' \left(u\right) = 5 {\left(y - 1\right)}^{4}$

$h = {\left({y}^{2} + 5 y\right)}^{6}$

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

$h ' \left(u \left(y\right)\right) = h ' \left(u\right) \left(u ' \left(y\right)\right)$

let $u \left(y\right) = {y}^{2} + 5 y , u ' \left(y\right) = 2 y + 5$

$h ' \left(u \left(y\right)\right) = h ' \left(u\right) \left(u ' \left(y\right)\right)$

$h ' \left(u \left(y\right)\right) = 6 {\left({y}^{2} + 5 y\right)}^{5} \left(2 y + 5\right)$

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

$f ' \left(y\right) = \frac{\left(5 {\left(y - 1\right)}^{4}\right) {\left({y}^{2} + 5 y\right)}^{6} - {\left(y - 1\right)}^{5} \left(6 {\left({y}^{2} + 5 y\right)}^{5} \left(2 y + 5\right)\right)}{{y}^{2} + 5 y} ^ 12$

$f ' \left(y\right) = \frac{\left(5 {\left(y - 1\right)}^{4}\right) \left({y}^{2} + 5 y\right) - {\left(y - 1\right)}^{5} \left(6 \left(2 y + 5\right)\right)}{{y}^{2} + 5 y} ^ 7$

$f ' \left(y\right) = \frac{\left(5 {\left(y - 1\right)}^{4}\right) \left({y}^{2} + 5 y\right) - {\left(y - 1\right)}^{5} \left(12 y + 30\right)}{{y}^{2} + 5 y} ^ 7$