Qualification > Math

C3 and C4 DOUBTS HERE!!!

<< < (8/37) > >>

S.M.A.T:

--- Quote from: M-H on September 13, 2010, 10:18:15 am ---Asif, how did reach the 4th step ???--as in, after x+.5 it becomes 2x+1..kibhabe??? :(

--- End quote ---

4((x+0.5)^2)
=4((x+(1/2))^2)
=4(((2x+1)/2)^2)
=4(((2x+1)^2)/(2^2))
=4(((2x+1)^2)/4)
=(2x+1)

If you don't understand then tell me :)

astarmathsandphysics:
I will look at that.

M-H:
Find dy/dx for-
x=1/(2t-1), y=(t^2)/(2t-1)

I don't know how to differentiate functions with reciprocals...

Thanks a bunch in advance.

iluvme:

--- Quote from: M-H on September 24, 2010, 03:17:12 pm ---Find dy/dx for-
x=1/(2t-1), y=(t^2)/(2t-1)

I don't know how to differentiate functions with reciprocals...

Thanks a bunch in advance.

--- End quote ---

i) x=1/(2t-1),
   dy/dx= 1(2t-1)-1
             =-1.(2t-1)-1-1.(2)
             =-1.(2t-1)-2.(2)
             =-2/(2t-1)2
 
ii) y=t2/(2t-1)
       I'm sure you know the UV method, therefore,
     
    dy/dx=2t(2t-1)-t2.(2)/(2t-1)2

S.M.A.T:

--- Quote from: M-H on September 24, 2010, 03:17:12 pm ---Find dy/dx for-
x=1/(2t-1), y=(t^2)/(2t-1)

I don't know how to differentiate functions with reciprocals...

Thanks a bunch in advance.

--- End quote ---

Here :)

Navigation

[0] Message Index

[#] Next page

[*] Previous page