Question

find the derivative of the function.

$y = \frac{5x - 9}{x^{2}+7x}$

$y = \square$

Explanation:

Step1: Apply quotient rule

Let \( u = 5x - 9\), \(u^\prime=5\); \(v=x^{2}+7x\), \(v^\prime = 2x + 7\). The quotient rule is \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\).
So \(y^\prime=\frac{5(x^{2}+7x)-(5x - 9)(2x + 7)}{(x^{2}+7x)^{2}}\).

Step2: Expand numerator

Expand \(5(x^{2}+7x)=5x^{2}+35x\) and \((5x - 9)(2x + 7)=10x^{2}+35x-18x - 63=10x^{2}+17x - 63\).
Then \(y^\prime=\frac{5x^{2}+35x-(10x^{2}+17x - 63)}{(x^{2}+7x)^{2}}\).

Step3: Simplify numerator

\(5x^{2}+35x - 10x^{2}-17x + 63=-5x^{2}+18x + 63\).
So \(y^\prime=\frac{-5x^{2}+18x + 63}{(x^{2}+7x)^{2}}\).

Answer:

\(\frac{-5x^{2}+18x + 63}{(x^{2}+7x)^{2}}\)