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y= 1/ (x+2)(x-3)
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
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
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 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin
dear vineeta
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
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin
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