Need help to find dx/dy

• Snoopyies

  18 Nov 10, 21:10

  Question Given that x = 3/(y-2), find dx/dy Thanks for the help.
• sbst275

  18 Nov 10, 21:15

  Originally posted by Snoopyies:

  Question

  Given that x = 3/(y-2), find dx/dy

  Thanks for the help.

  take out '3' as it's a constant.

  
  let u = y - 2

  du/dy = 1

  x = 1/ u

  3 dx/du = 3 ln u 

  therefore dx/ dy = 3 ln (y-2)

• sbst275

  18 Nov 10, 21:18

  use e chain rule

  dx/ dy = dx/ du du/dy

• Forbiddensinner

  18 Nov 10, 21:54

  Originally posted by sbst275:

  take out '3' as it's a constant.

  
  let u = y - 2

  du/dy = 1

  x = 1/ u

  3 dx/du = 3 ln u 

  therefore dx/ dy = 3 ln (y-2)

  I will suggest something else instead.

  x=3/(y-2)

  dx/dy

  = [-3/(y-2)^2][d/dy (y-2)] 

  = [-3/(y-2)^2](1)

  = -3/(y-2)^2

• eagle

  19 Nov 10, 10:59

  What level are you at?

  If A levels H2 Maths, one should be able to jump direct to the answer in one step.

  If O levels or H1 Maths, then the easiest wat is to change to 3(y-2)^-1, and differentiate using chain rule

• SBS n SMRT

  20 Nov 10, 23:13

  Originally posted by eagle:

  What level are you at?

  If A levels H2 Maths, one should be able to jump direct to the answer in one step.

  If O levels or H1 Maths, then the easiest wat is to change to 3(y-2)^-1, and differentiate using chain rule

  How to jump in H2 maths, using formulae? MF15?

• eagle

  22 Nov 10, 02:00

  It should already be familiar that differentiating 1/x w.r.t. x gives -1/x^2

  Hence, differentiating 3/(y-2) w.r.t. y will immediately gives us -3/(y-2)^2

   

  This is what I expect of my H2 maths students.