Question

let $y=\frac{x^{2}+2x}{4 - 5x}$. what is the value of $\frac{dy}{dx}$ at $x = 2$? choose 1 answer:

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$. Here, $u = x^{2}+2x$ and $v = 4 - 5x$. First, find $\frac{du}{dx}$ and $\frac{dv}{dx}$. $\frac{du}{dx}=2x + 2$ and $\frac{dv}{dx}=-5$.

Step2: Substitute into quotient - rule formula

$\frac{dy}{dx}=\frac{(4 - 5x)(2x + 2)-(x^{2}+2x)(-5)}{(4 - 5x)^{2}}$. Expand the numerator: $(4 - 5x)(2x + 2)=8x+8-10x^{2}-10x=-10x^{2}-2x + 8$, and $(x^{2}+2x)(-5)=-5x^{2}-10x$. Then $\frac{dy}{dx}=\frac{-10x^{2}-2x + 8 + 5x^{2}+10x}{(4 - 5x)^{2}}=\frac{-5x^{2}+8x + 8}{(4 - 5x)^{2}}$.

Step3: Evaluate at $x = 2$

Substitute $x = 2$ into $\frac{dy}{dx}$. $\frac{dy}{dx}\big|_{x = 2}=\frac{-5(2)^{2}+8(2)+8}{(4-5\times2)^{2}}=\frac{-20 + 16+8}{(4 - 10)^{2}}=\frac{4}{36}=\frac{1}{9}$.

Answer:

$\frac{1}{9}$