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y= 1/ (x+2)(x-3)

vineeta nadig , 15 Years ago
Grade 11
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JEEVANT MEHTA

Last Activity: 15 Years ago

logy= log1 - log(x+2) - log(x-3)

 1/y  (y')  = 0 - 1/(x+2) - 1/(x-3)

               = (-x+3-x-2)/(x+2)(x-3)

               = (-2x+1)/(x+2)(x-3)

 1/(x+2)(x-3)  y'    = (-2x+1)/(x+2)(x-3)

                       y'   = -2x+1 

Badiuddin askIITians.ismu Expert

Last Activity: 15 Years ago

dear vineeta

y= 1/ (x+2)(x-3)

or (x+2)(x-3) =1/y

    differtiate

  2x-1 =-1/y2 dy/dx

or dy/dx  =-y2 (2x-1)

                    =-(2x-1)/(x+2)2(x-3)2


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