Question

can sumone halp me please???http://prnt.sc/ckh6xh

Answer 1

\[\large\rm \frac{e^x(x^2+4)^{1/2}-e^x(3x)(3x^2+4)^{-1/2}}{(3x^2+4)}\]Ok so we took an e^x out of each term,\[\large\rm \frac{e^x\left[(x^2+4)^{1/2}-(3x)(3x^2+4)^{-1/2}\right]}{(3x^2+4)}\]And the other factoring step is confusing you?

Answer 2

if u could show me that step in pieces, i'd pay you......respect

Answer 3

yeah...i mean i didnt even know, it was factoring that was confusing me but ok...yeah

Answer 4

Ahh typo sorry D:\[\large\rm \frac{e^x\left[(3x^2+4)^{1/2}-(3x)(3x^2+4)^{-1/2}\right]}{(3x^2+4)}\]

Answer 5

plz make it easy tho....k?? thanks

Answer 6

So it looks like they factored (3x^2+4)^{-1/2} out of each term,\[\large\rm \frac{e^x(3x^2+4)^{-1/2}\left[\frac{(3x^2+4)^{1/2}}{(3x^2+4)^{-1/2}}-\frac{(3x)(3x^2+4)^{-1/2}}{(3x^2+4)^{-1/2}}\right]}{(3x^2+4)}\]It's clear what happens to the second term when we do this, yes?\[\large\rm \frac{e^x(3x^2+4)^{-1/2}\left[\frac{(3x^2+4)^{1/2}}{(3x^2+4)^{-1/2}}-3x\right]}{(3x^2+4)}\]

Answer 7

If I draw it, maybe easier.