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Suppose we wanted to differentiate the function

$$h(x) = (2-2x^3)^4 + \frac{1}{2-2x^3}$$

using the chain rule, writing the function as the composite $h(x) = f(g(x))$. Identify the functions $f(x)$ and $g(x)$, calculate the derivatives of these two functions and now calculate the derivatives of $h(x)$ using the chain rule.

Please help me solve this! I'm mostly having trouble rewriting the function as $h(x) = f(g(x))$...

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The Chain Rule in Newton's notation says If $h(x) = f(g(x))$, then $h'(x) = f'(g(x)) * g'(x)$

Since we see $2 - 2x^3$ come up in $h(x)$, that gives us a strong suggestion that we can use $g(x) = 2 - 2x^3$ and, therefore, $f(x) = x^4 + \frac{1}{x}$

What remains is to find $f'(x)$ and $g'(x)$, and then plug it into the formula.

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Hint: If $g(x) = 2 - 2x^3$, what's $f(x)?$

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If $u=2-2x^3$ then $h(x) = u^4+\dfrac 1 u$. So call that $f(u)$, and let $u=2-2x^3=g(x)$.

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Let $g(x)=2-2x^3$. Then, let $f(x)=x^4+\frac{1}{x}$.

The chain rule states $h^{'}(x)=f^{'}(g(x))g^{'}(x)$. Using $g^{'}(x)=-6x^2,f^{'}(x)=4x^3+-\frac{1}{x^2}$, you should be able to find out $h^{'}(x)$.

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