Maths P1-Help Needed!!

[ashutoshamatya]

8. A curve has equation y=x/(2*x^2+1)^0.5

(a) Show that dy/dx= (2*x^2+1)^(-3/2)
(b) Hence show that the curve has no turning point.

 

[Nibz]

y = x/(2x^2 +1)^1/2

dy/dx = [(2x^2 +1)^(1/2) - (x) (1/2 *(2x^2 +1)^(-1/2)*(4x)] /(2x^2+1)

[(2x^2 +1)^(1/2) - (4x^2) /2(2x^2 +1)^(1/2)] /(2x^2 +1)

Take L.C.M of the inside part!

We have => [(2(2x^2 +1) - 4x^2) /2(2x^2 +1)^(1/2)) ] /(2x^2 +1)

=> [ ( 1 /(2x^2 +1)^(1/2)) ] /(2x^2 +1)

=> 1/(2x^2 +1)^(1/2) . 1/(2x^2 +1) => 1/(2x^2 +1)^(3/2) => (2x^2 +1)^(-3/2) (SHOWN!)

Now (b) part => 1 /(2x^2 +1)^3/2 = 0
1=0 ... since the first derivative is not equal to zero.. thus the curve has no turning points!

Hope this helps!

 

[ahmed t]

isnt that the quotient rule which is not in AS

 

[ashutoshamatya]

Yepp.. it is the quotient rule..

 

[ahmed t]

so why write P1