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differentiate xy=ae^x+be^-x+x^2 where a,b are arbitary constants

differentiate xy=ae^x+be^-x+x^2 where a,b are arbitary constants

Grade:12th pass

1 Answers

Nandana
110 Points
7 years ago
It’s simple derivative of y  with respect to x
        which is given by -
y=(ae^x+be^-x+x^2)/x
       y = ae^x/x+be^-x/x+x
       by using the formula that d(u/v) = v du -u dv/v^2 , we obtain
        d/dx(y) =  a [xe^x- e^x /x^2 ] +b [-x e^-x – e^-x/x^2]+1
            simplify forther if you  want !

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