Differentiate equation with parenthesis I have a problem. I'm studying calculus, but I don't have a good math background, so I have a problem: I don't know well how to differentiate an equation with parenthesis.
The equation is the following:
$f(x) = 25x^3(x-1)^2$
Is it correct to use the Differentiation Product Rule in this way:
$f'(x)=75x^2*(x-1)^2+25x^3*2(x-1)$
or before I have to solve $(x-1)^2$ in this way:
f(x) = $25x^3*(x^2+1-2x)$ and then
     = $25x^5+25x^3-50x^4$
?
Thanks in advance
 A: 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.
A: 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: 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 $$\frac{f'(x)}{f(x)}=\frac 3 x+\frac{2}{x-1}=\frac{5x-3}{x(x-1)}$$ $$f'(x)=f(x)\frac{5x-3}{x(x-1)}=25x^3(x-1)^2\frac{5x-3}{x(x-1)}=25x^2(x-1)(5x-3)$$
A: Yes.  Just think of each sub-expression as a separate function, or possibly a composition of functions.  Then all derivatives can be resolved by application of the product and chain rules.
Example: Think of the sub-expression $x^3$ as the expanded product $x.x.x$.  Applying the product rule yields $x'.x.x + x.x'.x +x.x.x'$ = $1.x.x + x.1.x + x.x.1$ = ... (repacking) = $3x^2$.
Of course, once you hit $sin()$, $log()$, etc., you have to resort to something else.
