Question

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

Explanation:

Step1: Recall quotient - rule

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

Step2: Find $u^{\prime}(x)$ and $v^{\prime}(x)$

Differentiate $u(x)$ and $v(x)$ separately. For $u(x)=10x - 2$, $u^{\prime}(x)=10$ (using the power - rule $\frac{d}{dx}(ax + b)=a$). For $v(x)=4x^{2}$, $v^{\prime}(x)=8x$ (using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$).

Step3: Apply the quotient - rule

Substitute $u(x)$, $u^{\prime}(x)$, $v(x)$ and $v^{\prime}(x)$ into the quotient - rule formula:
\[

$$\begin{align*} h^{\prime}(x)&=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}\\ &=\frac{10\times(4x^{2})-(10x - 2)\times(8x)}{(4x^{2})^{2}}\\ &=\frac{40x^{2}-(80x^{2}-16x)}{16x^{4}}\\ &=\frac{40x^{2}-80x^{2}+16x}{16x^{4}}\\ &=\frac{- 40x^{2}+16x}{16x^{4}}\\ &=\frac{-40x^{2}}{16x^{4}}+\frac{16x}{16x^{4}}\\ &=-\frac{5}{2x^{2}}+\frac{1}{x^{3}} \end{align*}$$

\]

Answer:

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