Question

hw14 the chain rule (target c4; §3.6)
score: 1/11 answered: 1/11
question 2
use the chain rule to find the derivative of
$f(x)=8sqrt{3x^{4}+2x^{8}}$
type your answer without fractional or negative exponents.
$f(x)=$
question help: video message instructor
submit question

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=8\sqrt{3x^{4}+2x^{8}}$ as $f(x)=8(3x^{4}+2x^{8})^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Let $u = 3x^{4}+2x^{8}$, so $y = 8u^{\frac{1}{2}}$. First, find the derivative of $y$ with respect to $u$: $\frac{dy}{du}=8\times\frac{1}{2}u^{-\frac{1}{2}} = 4u^{-\frac{1}{2}}$. Then find the derivative of $u$ with respect to $x$: $\frac{du}{dx}=12x^{3}+16x^{7}$.

Step3: Calculate $f'(x)$

By the chain - rule $f'(x)=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $u = 3x^{4}+2x^{8}$ back into $\frac{dy}{du}$ and multiply by $\frac{du}{dx}$:
\[

$$\begin{align*} f'(x)&=4(3x^{4}+2x^{8})^{-\frac{1}{2}}\cdot(12x^{3}+16x^{7})\\ &=\frac{4(12x^{3}+16x^{7})}{\sqrt{3x^{4}+2x^{8}}}\\ &=\frac{48x^{3}+64x^{7}}{\sqrt{3x^{4}+2x^{8}}}\\ &=\frac{48x^{3}+64x^{7}}{\sqrt{x^{4}(3 + 2x^{4})}}\\ &=\frac{48x^{3}+64x^{7}}{x^{2}\sqrt{3 + 2x^{4}}}\\ &=\frac{48x+64x^{5}}{\sqrt{3 + 2x^{4}}} \end{align*}$$

\]
Multiply the numerator and denominator by $\sqrt{3 + 2x^{4}}$ to get rid of the square - root in the denominator:
\[

$$\begin{align*} f'(x)&=\frac{(48x + 64x^{5})\sqrt{3+2x^{4}}}{3 + 2x^{4}}\\ &=\frac{48x\sqrt{3 + 2x^{4}}+64x^{5}\sqrt{3 + 2x^{4}}}{3 + 2x^{4}} \end{align*}$$

\]

Answer:

$\frac{48x\sqrt{3 + 2x^{4}}+64x^{5}\sqrt{3 + 2x^{4}}}{3 + 2x^{4}}$