question use the quotient rule to find the derivative h(x) of the following function. h(x)=\frac{4x^{2}}{10x…

Explanation:

Step1: Recall quotient - rule formula

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

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

Differentiate $u(x)$: $u'(x)=\frac{d}{dx}(10x - 2)=10$. Differentiate $v(x)$: $v'(x)=\frac{d}{dx}(4x^{2})=8x$.

Step3: Apply the quotient - rule

Substitute $u(x),u'(x),v(x),v'(x)$ into the quotient - rule formula. [ \begin{align*} h'(x)&=\frac{10\times(4x^{2})-(10x - 2)\times(8x)}{(4x^{2})^{2}}\ &=\frac{40x^{2}-(80x^{2}-16x)}{16x^{4}}\ &=\frac{40x^{2}-80x^{2}+16x}{16x^{4}}\ &=\frac{-40x^{2}+16x}{16x^{4}}\ &=\frac{-40x^{2}}{16x^{4}}+\frac{16x}{16x^{4}}\ &=-\frac{5}{2x^{2}}+\frac{1}{x^{3}} \end{align*} ]

Answer:

$h'(x)=-\frac{5}{2x^{2}}+\frac{1}{x^{3}}$