Finding the Stationary Point of a Curve: y = 4 - 2/ (2x-1)^2 -x

[ibysaiyan]

Homework Statement

Hi all , again i got stuck onto this iffy bit , question is:
find dy/dx of the curve and hence its coordinates of the statinonary point.
y= 4 - 2/(2x-1)^2 -x

Homework Equations

du/dx x dy/du

The Attempt at a Solution

well this is how i started off:
let u = (2x-1) , y= 4-2(u)^-2 - x
du/dx= 2;dy/du:4u^-3-1

2x[4u^(-3)-1]
[8u^(-3) -2]
to find stationary points i will let dy/dx= 0 and solve the equation ?am i on the right track so far?

 

[ibysaiyan]

oo i think i have made a mistake.. i should have multiplied gradient u (2) with the first term for the gradient of dy/du ?
EDIT:sorry for being such a spammer,i managed to solve this question but grr.. i am stuck on the next one.
question is:
curve y = (3x-2)^3 - 5x^2
find dy/dx and stationary points.

My attempt :
dy/dx= 9u^2 - 10x
we know u= (3x-2), so equation becomes
9(3x-2)^2- 10x
81x^2 -108x +26= 0
divide by 9
9x^2 -12x -2.8 ( am i over-complicating it?)

 

[HallsofIvy]

oo i think i have made a mistake.. i should have multiplied gradient u (2) with the first term for the gradient of dy/du ?
EDIT:sorry for being such a spammer,i managed to solve this question but grr.. i am stuck on the next one.
question is:
curve y = (3x-2)^3 - 5x^2
find dy/dx and stationary points.

My attempt :
dy/dx= 9u^2 - 10x
we know u= (3x-2), so equation becomes
9(3x-2)^2- 10x
81x^2 -108x +26= 0

Check your arithmetic! 9(9x^2- 12x+ 4)- 10x= 81x^2- 108x+ 36- 10x.

divide by 9
9x^2 -12x -2.8 ( am i over-complicating it?)