Maths Paper 1 Diffrenciation Question

[allysaleemally]

im stuck at paper 1 may june 2005 question 2
Paper: http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s05_qp_1.pdf
MS: http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s05_ms_1.pdf
could you please explain it to me
i dont understand in the marksheet how '-12(x2 - 4x)to the power -2'

 

[Khan_971]

Assalam walaikum!
Well, In differentiation of a fraction, the rule is:
if y=u/v, then dy/dx =
[u * d/dx (v) - v* d/dx (u) ] / v^2

E.g, y= 1/x
dy/dx = [1 (1) - x (0)] / x^2 [d/dx (x) = 1, d/dx (any constant, i.e, 0,1,2,3....) =0]
therefore dy/dx = 1/x^2

Since the equation is 12/(x^2)-4x

dy/dx = [ {(x^2)-4x} * d/dx (12) - 12* d/dx {(x^2)-4x} ] / {(x^2)-4x}^2

dy/dx = -12 (2x-4) {which is d/dx of (x^2)-4x} / {(x^2)-4x}^2 d/dx of (12) = 0

when you bring the denominator upwards, its power always becomes of the reverse sign.

Which in this case will become negative, (Inversing the denominator will give a -ve sign to the power it has)
Hope this solves the problem.
If you do not understand this, let me know so I can attach a page in normal handwriting.

 

[Nikesh]

here is the solution......
hope you get it