Question

webwork 2 - topics 4 (1 point) if $f(x)=8x^{2}arctan(4x^{2})$, find $f(x)$. $f(x)=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = 8x^{2}$ and $v=\arctan(4x^{2})$. First, find $u'$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $u'=\frac{d}{dx}(8x^{2})=16x$.

Step2: Apply chain - rule to find $v'$

The derivative of $\arctan(t)$ is $\frac{1}{1 + t^{2}}$. Let $t = 4x^{2}$. Then, by the chain - rule $\frac{d}{dx}(v)=\frac{d}{dt}(\arctan(t))\cdot\frac{dt}{dx}$. We know that $\frac{d}{dt}(\arctan(t))=\frac{1}{1 + t^{2}}$ and $\frac{dt}{dx}=\frac{d}{dx}(4x^{2}) = 8x$. So $v'=\frac{8x}{1+(4x^{2})^{2}}=\frac{8x}{1 + 16x^{4}}$.

Step3: Calculate $f'(x)$

Using the product - rule $f'(x)=u'v+uv'$, we substitute $u = 8x^{2}$, $u' = 16x$, $v=\arctan(4x^{2})$, and $v'=\frac{8x}{1 + 16x^{4}}$ into the formula:
\[

$$\begin{align*} f'(x)&=16x\arctan(4x^{2})+8x^{2}\cdot\frac{8x}{1 + 16x^{4}}\\ &=16x\arctan(4x^{2})+\frac{64x^{3}}{1 + 16x^{4}} \end{align*}$$

\]

Answer:

$16x\arctan(4x^{2})+\frac{64x^{3}}{1 + 16x^{4}}$