Hello. In a few hours I have to take a test concerning derivatives, I understand almost everything, however I'm having a hard time trying to understand an exercise which is already solved in the book, this is the excercise

find f'(x) if f(x) = (2x^2+3)^4 (3x-1)^5

and so they solve the excercise, they apply product rule followed by the general power rule, but I'm so confused, I'm lost, I don't get it, i don't understand how they solve that excercise.

Could you please explain to me how to solve an exercise of that nature? please, with easy steps.
thank you.

2. If you let $\displaystyle (2x^2+3)^4 = A$ and $\displaystyle (3x-1)^5 = B$
You can see that the product rule will yield the derivative: $\displaystyle \frac{dA}{dx}B + \frac{dB}{dx}A$

With this in hand, you must find the derivatives of A and B.

To do this you can use the chain rule: $\displaystyle \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$

For the example of A, let $\displaystyle 2x^2+3 = u$

You can combine all of this into one formula, but it's easiest done in seperate steps.