Question

use the quotient rule to find the derivative $h(x)$ of the following function. $h(x)=\frac{4x^{2}}{10x - 2}$. provide your answer below: $h(x)=square$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $h(x)=\frac{u(x)}{v(x)}$, then $h'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}$. Here, $u(x)=10x - 2$ and $v(x)=4x^{2}$.

Step2: Find $u'(x)$ and $v'(x)$

Differentiate $u(x)=10x - 2$ using the power - rule. The derivative of $ax + b$ is $a$, so $u'(x)=10$. Differentiate $v(x)=4x^{2}$ using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, so $v'(x)=8x$.

Step3: Apply the quotient - rule

$h'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}=\frac{10\times4x^{2}-(10x - 2)\times8x}{(4x^{2})^{2}}$.

Step4: Expand and simplify

First, expand the numerator: $10\times4x^{2}-(10x - 2)\times8x = 40x^{2}-(80x^{2}-16x)=40x^{2}-80x^{2}+16x=-40x^{2}+16x$. The denominator is $(4x^{2})^{2}=16x^{4}$. So $h'(x)=\frac{-40x^{2}+16x}{16x^{4}}=\frac{-40x^{2}}{16x^{4}}+\frac{16x}{16x^{4}}=-\frac{5}{2x^{2}}+\frac{1}{x^{3}}$.

Answer:

$-\frac{5}{2x^{2}}+\frac{1}{x^{3}}$