Question

find f'x when f(x)= (2x^7-x^2)(x-1/x+1)

Answer 1

:o

Answer 2

\[\Large\bf f(x)=(2x^7-x^2)\left(\frac{x-1}{x+1}\right)\]
\[\Large\bf f'(x)=\color{#DD4747}{\left(2x^7-x^2\right)'}\left(\frac{x-1}{x+1}\right)+(2x^7-x^2)\color{#DD4747}{\left(\frac{x-1}{x+1}\right)'}\]

Answer 3

So we start with product rule

Answer 4

okay i did that lol

Answer 5

So we have to take the derivative of the red parts.
Can you do the first one?

Answer 6

yes i can , (14x^7-2x) and (2x/(x+1)^2)?

Answer 7

14x^6-2x

Answer 8

Applying Quotient rule:
\[\Large\bf \color{#DD4747}{\left(\frac{x-1}{x+1}\right)'}\quad=\quad \frac{(1)(x+1)-(x-1)(1)}{(x+1)^2}\]

Hmm I think your numerator is off, our x's will cancel out, yes?

Answer 9

\[\Large \bf =\frac{2}{(x+1)^2}\]

Answer 10

oh okay yeah it was lol

Answer 11

so after that do we find a common denominator?

Answer 12

So we're here?\[\large\bf f'(x)=\color{royalblue}{\left(14x^6-2x\right)}\left(\frac{x-1}{x+1}\right)+(2x^7-x^2)\color{royalblue}{\left(\frac{2}{(x+1)^2}\right)}\]

Answer 13

Mmmm yah I guess you could get a common denominator if you want.
It's best to just stop here though.

If your teacher wants it simplified further though,
yes we would want a common denominator of (x+1)^2.
So the first term needs another factor of (x+1) in the top and bottom.

Answer 14

i think ill stop here because itll be too much work lol

Answer 15

XD good idea haha