√x−1⋅(1−6x)−12√x−1⋅(x−3x2)x−1
=[√x−1⋅(1−6x)−12⋅√x−1⋅(x−3x2)x−1]⋅2⋅√x−12⋅√x−1=√x−1⋅2⋅√x−1⋅(1−6x)−2⋅√x−12⋅√x−1⋅(x−3x2)2⋅(x−1)⋅√x−1=2⋅(√x−1)2⋅(1−6x)−(x−3x2)2⋅(x−1)⋅√x−1=2⋅(x−1)⋅(1−6x)−(x−3x2)2⋅(x−1)⋅√x−1=2⋅(x−1)⋅(1−6x)−x+3x22⋅(x−1)⋅√x−1=2⋅(x−6x2−1+6x)−x+3x22⋅(x−1)⋅√x−1=2⋅(7x−6x2−1)−x+3x22⋅(x−1)⋅√x−1=14x−12x2−2−x+3x22⋅(x−1)⋅√x−1=13x−9x2−22⋅(x−1)⋅√x−1=−9x2+13x−22⋅(x−1)⋅√x−1=−9x2+13x−22⋅(x−1)⋅(x−1)12=−9x2+13x−22⋅(x−1)32
Hi Namodesto,
You cannot solve it because it is not an equation.
You can only simplify it
√x−1∗(1−6x)−(1/√x−1∗2)∗(x−3x2)x−1=[√x−1∗(1−6x)−(2√x−1)∗(x−3x2)]÷(x−1)=[(x−1)(1−6x)−2(x−3x2)√x−1]÷(x−1)=[(x−1)(1−6x)−2(x−3x2)√x−1]×1(x−1)=[(x−1)(1−6x)−2(x−3x2)√x−1]×1(x−1)=x−6x2−1+6x−2x+6x2√x−1∗(x−1)=5x−1(x−1)3/2
Hi Melody,
http://math.stackexchange.com/questions/467592/derivative-of-frac-x-cdot-left1-3x-right-sqrtx-1?rq=1
I have problem understanding the last step. The following problem above was taken from an example of this website. How did he get to the final answer being -9x^2 + 13x - 2 from what I've written above?
Hi again Namodesto,
A part of the problem is that you presented the question incorrectly. 1/(sqrt(x-1)*2) the bottom needed to be in brackets.
You did not hve brackets which meant your 2 was up the top when it should have been down the bottom.
I have been called away. I will be back in about a half hour. sorry.
I got t there answer except I got -2 anstead of +2
I will look more in a while :))
Must run .
Ok I am back and I can see that Heureka is already preparing another answer for you.
That +2 on the first answer on this web page that you have since inserted :-
http://math.stackexchange.com/questions/467592/derivative-of-frac-x-cdot-left1-3x-right-sqrtx-1?rq=1
is a typo, the answer further down is -2 and it is the same as I got when I redid your expression with the 2 down the bottom instead of up the top. :)
I'll leave it now untill we both see what Heureka is preparing for you :))
√x−1⋅(1−6x)−12√x−1⋅(x−3x2)x−1
=[√x−1⋅(1−6x)−12⋅√x−1⋅(x−3x2)x−1]⋅2⋅√x−12⋅√x−1=√x−1⋅2⋅√x−1⋅(1−6x)−2⋅√x−12⋅√x−1⋅(x−3x2)2⋅(x−1)⋅√x−1=2⋅(√x−1)2⋅(1−6x)−(x−3x2)2⋅(x−1)⋅√x−1=2⋅(x−1)⋅(1−6x)−(x−3x2)2⋅(x−1)⋅√x−1=2⋅(x−1)⋅(1−6x)−x+3x22⋅(x−1)⋅√x−1=2⋅(x−6x2−1+6x)−x+3x22⋅(x−1)⋅√x−1=2⋅(7x−6x2−1)−x+3x22⋅(x−1)⋅√x−1=14x−12x2−2−x+3x22⋅(x−1)⋅√x−1=13x−9x2−22⋅(x−1)⋅√x−1=−9x2+13x−22⋅(x−1)⋅√x−1=−9x2+13x−22⋅(x−1)⋅(x−1)12=−9x2+13x−22⋅(x−1)32