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
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 $$\frac{f'(x)}{f(x)}=\frac 3 x+\frac{2}{x-1}=\frac{5x-3}{x(x-1)}$$ $$f'(x)=f(x)\frac{5x-3}{x(x-1)}=25x^3(x-1)^2\frac{5x-3}{x(x-1)}=25x^2(x-1)(5x-3)$$
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.
