Question

find dy/dx if y=((x^5+2x^4-3x^3+2x-1)^500)/(x)

Answer 1

Well you first need to expand out the 500th degree term, obviously.
They divide each of those 501 terms by x.
Then apply the power rule to each term.
obvs.

Answer 2

ok ok jk jk...
Sooo uh, quotient rule to start, yes?

Answer 3

@zepdrix right...so I have 500(5x^4+8x^3-9x^2+2x^499+dy/dx(x) Correct?

Answer 4

Hmm i dunno what that is :o\[\Large\sf y=\frac{(x^5+2x^4-3x^3+2x-1)^{500}}{x}\]Hmm this is going to be difficult to squeeze on one line, let's see if I can do it.\[\sf y=\frac{\color{royalblue}{[(x^5+2x^4-3x^3+2x-1)^{500}]'}x-(x^5+2x^4-3x^3+2x-1)^{500}\color{royalblue}{[x]'}}{x^2}\]That would be our setup for the quotient rule.
(Dtop)Bottom minus Top(Dbottom) divided by bottom squared.
We need to take the derivative of the blue parts.

Answer 5

Derivative of that first term is a little tricky.
First, applying the power rule to the outside, good.\[\large\sf 500(x^5+2x^4-3x^3+2x-1)^{499}\color{royalblue}{(x^5+2x^4-3x^3+2x-1)'}\]We also must apply the chain rule,
multiplying by the derivative of the inner function.

Answer 6

Alternatively, you can rewrite the original equation so that differentiating would be a (imo) simpler process with the product rule:
\[y=\left(x^5+2x^4-3x^3+2x-1\right)^{500}x^{-1}\]
Then,
\[\frac{dy}{dx}=\frac{d}{dx}\left[\left(x^5+2x^4-3x^3+2x-1\right)^{500}\right]x^{-1}+\left(x^5+2x^4-3x^3+2x-1\right)^{500}\frac{d}{dx}\left[x^{-1}\right]\]

Answer 7

cutoff part: \(\dfrac{d}{dx}[x^{-1}]\).