Derivative of $(2x^2+3)^2(x^3-2x)^4$ I want to know if there's another method in shorter way. I came up to this problem

Find the $f'(x)$
$$
f(x) = (2x^2+3)^2\cdot(x^3-2x)^4
$$
Applying product rules
$$
\frac{d}{dx}f(x)g(x) = f(x)\frac{d}{dx}g'(x) + g'(x)\frac{d}{dx}f(x)
$$

Solve:
$$ f = (2x^2+3)^2 $$
$$ g = (x^3-2x)^4$$
$$
\frac{d}{dx}(2x^2+3)^2(x^3+2x)^4 + \frac{d}{dx}(x^3-2x)^4(2x^2+3)^2
$$
$$
2(2x^2+3) 4x(x^3-2x)^4 + 4(x^3-2x)^3(3x^2-2)(2x^2+3)^2
$$
$$
8x(2x^2+3)(x^3-2x)^4 + 4(x^3-2x)^3(3x^2-2)(2x^2+3)^2
$$
$$
8x(2x^2+3)(x^3-2x)^4 + (x^3-2x)^3(12x^2-8)(2x^2+3)^2
$$
Answer:
$$
f'(x) =8x(2x+3)(x^3-2x)^4 + (x^3-2x)^3(12x^2-8)(2x^2+3)^2
$$
Is this correct way or are there other methods?
Thanks in advance.
 A: Your method is correct, and since the question states that you should make use of the product rule, it is the right solution.
If you're new to derivatives, you may not yet have learned about derivatives of logs or about implicit differentiation; however you're clearly already familiar with the chain rule, so you should be learning about these other topics very soon.
Once you do, there is a more efficient approach which takes advantage of the properties of logs:
$$\begin{align}
f(x) &= (2x^2+3)^2\cdot(x^3-2x)^4 \\[12pt]
\ln f(x) &= \ln \left[ (2x^2+3)^2\cdot(x^3-2x)^4 \right] \\
&= 2 \ln (2x^2+3) + 4 \ln (x^3-2x) \\[12pt]
\frac{f'(x)}{f(x)} &= 2 \left( \frac{4x}{2x^2+3} \right) + 4 \left( \frac{3x^2-2}{x^3-2x} \right) \\[12pt]
f'(x) &= f(x) \left[ \frac{8x}{2x^2+3} + \frac{12x^2-8}{x^3-2x} \right] \\
&= 8x (2x^2+3) (x^3-2x)^4 + (12x^2-8) (2x^2+3)^2 (x^3-2x)^3
\end{align}$$
As @RyszardSzwarc notes in the comments below, technically the above method would not work when $x = 0, \pm \sqrt{2}$ as the function value would be zero, but that ultimately does not affect the solution.
