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:

$$
LATEXBLOCK0
$$

Answer:

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