Differentiate equation with parenthesis

Either way works. If you multiply out the products in your first result, you should discover that the two expressions you get for the derivative always have the same value.

Yes, they're both correct and equivalent. I'd use the first method and then factor out a $25x^2(x-1)$ to get $$f'(x) = 25x^2(x-1)\big(3(x-1) + 2x\big) = 25x^2(x-1)\big(5x - 3\big)$$

for the sake of neatness. If you multiply that out, you'll find that you get the same thing as expanding first and then differentiating term-wise.

A small (useful) trick when you face products, quotients, powers,.. : logarithmic differentiation.

Let us take your cas $$f(x) = 25x^3(x-1)^2\implies \log(f(x))=\log(25)+3\log(x)+2\log(x-1)$$ Now, differentiate $$\frac{f'(x)}{f(x)}=\frac 3 x+\frac{2}{x-1}=\frac{5x-3}{x(x-1)}$$ $$f'(x)=f(x)\frac{5x-3}{x(x-1)}=25x^3(x-1)^2\frac{5x-3}{x(x-1)}=25x^2(x-1)(5x-3)$$