Question

attempt 1: 10 attempts remaining. if $f(x)=\frac{x^{3}-7x}{8x^{2}+10}$, find $f(x)$. $f(x)=$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x^{3}-7x$ and $v = 8x^{2}+10$.

Step2: Find $u'$

Differentiate $u=x^{3}-7x$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $u'=\frac{d}{dx}(x^{3}-7x)=3x^{2}-7$.

Step3: Find $v'$

Differentiate $v = 8x^{2}+10$ with respect to $x$. Using the power - rule, $v'=\frac{d}{dx}(8x^{2}+10)=16x$.

Step4: Apply quotient - rule

Substitute $u$, $u'$, $v$, and $v'$ into the quotient - rule formula:
\[

$$\begin{align*} f'(x)&=\frac{(3x^{2}-7)(8x^{2}+10)-(x^{3}-7x)(16x)}{(8x^{2}+10)^{2}}\\ &=\frac{24x^{4}+30x^{2}-56x^{2}-70-(16x^{4}-112x^{2})}{(8x^{2}+10)^{2}}\\ &=\frac{24x^{4}+30x^{2}-56x^{2}-70 - 16x^{4}+112x^{2}}{(8x^{2}+10)^{2}}\\ &=\frac{(24x^{4}-16x^{4})+(30x^{2}-56x^{2}+112x^{2})-70}{(8x^{2}+10)^{2}}\\ &=\frac{8x^{4}+86x^{2}-70}{(8x^{2}+10)^{2}} \end{align*}$$

\]

Answer:

$\frac{8x^{4}+86x^{2}-70}{(8x^{2}+10)^{2}}$