Finding an expression for the derivative of $h(x)$ 
$f(x)$ can be differentiated.
  $$h(x) = ([f(x)]^2 + x^4)(7x-3)$$

I tried to solve this using the product rule:
$$ a(x) = ([f(x)]^2 + x^4)$$
$$ b(x) = (7x-3)$$
$$h'(x) = a'(x)\cdot b(x) + a(x)\cdot b'(x)$$
Not sure how to do $a'(x)$
 A: $$h(x)=([f(x)]^2+x^4)(7x−3)$$
$$ h'(x) = (2f(x)\cdot f'(x) + 4x^3)(7x-3) + ([f(x)]^2 + x^4)\cdot(7)$$
A: Just for the fun of the exercise, let us do it without using  chain and product rules.
Consider first $$h = (g + x^4)\,(7x-3)$$ Take logarithms $$\log(h)=\log(g+x^4)+\log(7x-3)$$ Differentiate both sides $$\frac{h'}h=\frac{g'+4x^3}{g+x^4}+\frac 7{7x-3}$$ Now, doing the same $$g=f^2\implies\log(g)=2\log(f)\implies\frac{g'}g=2\frac{f'}f\implies g'=2g\frac{f'}f=2f f'$$ This makes $$\frac{h'}h=\frac{2ff'+4x^3}{f^2+x^4}+\frac 7{7x-3}$$ $$h'=(f^2 + x^4)\,(7x-3)\left(\frac{2ff'+4x^3}{f^2+x^4}+\frac 7{7x-3} \right)$$ $$h'=(7x-3)(2ff'+4x^3)+7(f^2+x^4)$$
For sure, we could have also use the logarithmic differentiation for $x^4$ itself.
A: The explicit form of the chain rule for a function $g(x)$ is $$\frac{dg}{dx}=\frac{dg}{df}\frac{df}{dx}$$ where $f=f(x)$. In your case, $g(x)=[f(x)]^2$. So to find its derivative with respect to $x$, you just need to find it derivative with respect to $f(x)$ and the derivative of $f(x)$ with respect to $x$. 
For more information on this, I'll include a link to a page that gives a basic description of this use of chain rule. https://www.mathsisfun.com/calculus/implicit-differentiation.html
