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`        y= 1/ (x+2)(x-3)`
8 years ago

JEEVANT MEHTA
19 Points
```										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
```
8 years ago
147 Points
```										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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```
8 years ago
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