Check your arithmetic! 9(9x^2- 12x+ 4)- 10x= 81x^2- 108x+ 36- 10x.ibysaiyan said: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?)
A stationary point, also known as a critical point, is a point on a curve where the slope (or derivative) of the curve is zero. This means that the tangent line to the curve at that point is horizontal.
To find the stationary points of a curve, you must first find the derivative of the curve. Then, set the derivative equal to zero and solve for the x-value(s). These x-values will be the coordinates of the stationary points.
The derivative of the given curve, y = 4 - 2/ (2x-1)^2 -x, is y' = 4(2x-1)^-3 - 1. This can be simplified to y' = -8/(2x-1)^3 - 1.
The stationary points of a curve can be interpreted as the points where the curve changes direction, from increasing to decreasing or vice versa. They can also represent the maximum or minimum values of the curve.
A curve can have multiple stationary points, or it can have none. It is also possible for a curve to have a single stationary point that represents both a maximum and a minimum value.