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Grade 11Differential Calculus

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

vineeta nadig

16 Years agoGrade 11

2 Answers

JEEVANT MEHTA

Approved Tutor Answer16 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

16 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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Badiuddin

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