\(\dfrac{ \sqrt{x-1}\cdot(1-6x)- \frac{1}{2\sqrt{x-1}}\cdot (x-3x^2) } { x-1 }\)
\(\small{ \begin{array}{rcll} &=& \left[ \dfrac{ \sqrt{x-1}\cdot(1-6x)- \frac{1}{2\cdot \sqrt{x-1}}\cdot (x-3x^2) } { x-1 } \right] \cdot \dfrac{ 2\cdot \sqrt{x-1} } { 2\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ \sqrt{x-1}\cdot 2\cdot \sqrt{x-1}\cdot(1-6x)- \frac{2\cdot \sqrt{x-1}}{2\cdot \sqrt{x-1}}\cdot (x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot \left(\sqrt{x-1}\right)^2 \cdot(1-6x)-(x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot (x-1) \cdot(1-6x)- (x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot (x-1) \cdot(1-6x)- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot ( x-6x^2-1+6x )- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot ( 7x-6x^2-1 )- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 14x-12x^2-2 - x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 13x-9x^2-2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot (x-1)\cdot (x-1)^{\frac12} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot(x-1)^{\frac32} } \\\\ \end{array} }\)
.Hi Namodesto,
You cannot solve it because it is not an equation.
You can only simplify it
\({ \sqrt{x-1}*(1-6x)-(1/ \sqrt{x-1}*2)*(x-3x^2) \over x-1}\\ =\left[\sqrt{x-1}*(1-6x)-(\frac{2}{ \sqrt{x-1}})*(x-3x^2) \right]\div (x-1)\\ =\left[\frac{(x-1)(1-6x)-2 (x-3x^2)}{\sqrt{x-1}} \right]\div (x-1)\\ =\left[\frac{(x-1)(1-6x)-2 (x-3x^2)}{\sqrt{x-1}} \right]\times \frac{1}{ (x-1)}\\ =\left[\frac{(x-1)(1-6x)-2 (x-3x^2)}{\sqrt{x-1}} \right]\times \frac{1}{ (x-1)}\\ =\frac{x-6x^2-1+6x-2 x+6x^2}{\sqrt{x-1}*(x-1)} \\ =\frac{5x-1}{(x-1)^{3/2}} \\ \)
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 :))
\(\dfrac{ \sqrt{x-1}\cdot(1-6x)- \frac{1}{2\sqrt{x-1}}\cdot (x-3x^2) } { x-1 }\)
\(\small{ \begin{array}{rcll} &=& \left[ \dfrac{ \sqrt{x-1}\cdot(1-6x)- \frac{1}{2\cdot \sqrt{x-1}}\cdot (x-3x^2) } { x-1 } \right] \cdot \dfrac{ 2\cdot \sqrt{x-1} } { 2\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ \sqrt{x-1}\cdot 2\cdot \sqrt{x-1}\cdot(1-6x)- \frac{2\cdot \sqrt{x-1}}{2\cdot \sqrt{x-1}}\cdot (x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot \left(\sqrt{x-1}\right)^2 \cdot(1-6x)-(x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot (x-1) \cdot(1-6x)- (x-3x^2) } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot (x-1) \cdot(1-6x)- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot ( x-6x^2-1+6x )- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 2\cdot ( 7x-6x^2-1 )- x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 14x-12x^2-2 - x + 3x^2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ 13x-9x^2-2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot (x-1)\cdot \sqrt{x-1} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot (x-1)\cdot (x-1)^{\frac12} } \\\\ &=& \dfrac{ -9x^2+13x-2 } { 2\cdot(x-1)^{\frac32} } \\\\ \end{array} }\)