Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
y= 1/ (x+2)(x-3) y= 1/ (x+2)(x-3)
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
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -