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[Post New]29 Mar 2008 10:57:07 IST
    Subject: Re:Differentiation...
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ramkumar_november (1270)

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Olaaa!! Perrrfect answer. 230  [290 rates]
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y\;=\;sinx\;+\;e^x
 
\frac{dy}{dx}\;=\;cosx\;+\;e^x
 
\frac{dx}{dy}\;=\;\frac{1}{cosx\;+\;e^x}  ..................(1)
 
\frac{d^2x}{dy^2}\;=\;\bigg(\frac{-1}{(cosx+e^x)^2}\bigg).(e^x-sinx).\frac{dx}{dy}
 
\frac{d^2x}{dy^2}\;=\frac{sinx-e^x}{(cosx+e^x)^3}      from(1)
 
so 3) is the right option......
 
 
 this reply: 5 points  (with 1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
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