Question

hw14 the chain rule (target c4; 53.6)
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use the chain rule to find the derivative of
$f(x)=8sqrt{3x^{4}+2x^{8}}$
type your answer without fractional or negative exponents.
$f(x)=\frac{48xsqrt{3 + 2x^{4}}+64x^{7}sqrt{3 + 2x^{4}}}{3 + 2x^{4}}$
basic funcs trig
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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 $\frac{dy}{du}$ and $\frac{du}{dx}$.
$\frac{dy}{du}=8\times\frac{1}{2}u^{-\frac{1}{2}} = 4u^{-\frac{1}{2}}$.
$\frac{du}{dx}=(3x^{4}+2x^{8})'=3\times4x^{3}+2\times8x^{7}=12x^{3}+16x^{7}$.

Step3: Multiply $\frac{dy}{du}$ and $\frac{du}{dx}$

$f'(x)=\frac{dy}{du}\cdot\frac{du}{dx}=4u^{-\frac{1}{2}}\cdot(12x^{3}+16x^{7})$.
Substitute $u = 3x^{4}+2x^{8}$ back in:
$f'(x)=4(3x^{4}+2x^{8})^{-\frac{1}{2}}\cdot(12x^{3}+16x^{7})$.
$=\frac{4(12x^{3}+16x^{7})}{\sqrt{3x^{4}+2x^{8}}}=\frac{48x^{3}+64x^{7}}{\sqrt{3x^{4}+2x^{8}}}=\frac{48x\sqrt{3 + 2x^{4}}+64x^{5}\sqrt{3 + 2x^{4}}}{3 + 2x^{4}}$

Answer:

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