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Question-10408




Question Number 10408 by konen last updated on 07/Feb/17
Answered by mrW1 last updated on 08/Feb/17
y^2 =e^(x+y)   ⇒2ln y=x+y  ⇒x=2ln y−y  (dx/dy)=(2/y)−1=((2−y)/y)    (dy/dx)=(1/(dx/dy))=(1/((2−y)/y))=(y/(2−y))    let g=(dy/dx)=(y/(2−y))  (dg/dy)=(1/(2−y))+(y/((2−y)^2 ))=(2/((2−y)^2 ))    (d^2 y/dx^2 )=((d((dy/dx)))/dx)=(dg/dx)=(dg/dy)×(dy/dx)=(2/((2−y)^2 ))×(y/(2−y))=((2y)/((2−y)^3 ))    ⇒answer D
$${y}^{\mathrm{2}} ={e}^{{x}+{y}} \\ $$$$\Rightarrow\mathrm{2ln}\:{y}={x}+{y} \\ $$$$\Rightarrow{x}=\mathrm{2ln}\:{y}−{y} \\ $$$$\frac{{dx}}{{dy}}=\frac{\mathrm{2}}{{y}}−\mathrm{1}=\frac{\mathrm{2}−{y}}{{y}} \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\frac{{dx}}{{dy}}}=\frac{\mathrm{1}}{\frac{\mathrm{2}−{y}}{{y}}}=\frac{{y}}{\mathrm{2}−{y}} \\ $$$$ \\ $$$${let}\:{g}=\frac{{dy}}{{dx}}=\frac{{y}}{\mathrm{2}−{y}} \\ $$$$\frac{{dg}}{{dy}}=\frac{\mathrm{1}}{\mathrm{2}−{y}}+\frac{{y}}{\left(\mathrm{2}−{y}\right)^{\mathrm{2}} }=\frac{\mathrm{2}}{\left(\mathrm{2}−{y}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{d}\left(\frac{{dy}}{{dx}}\right)}{{dx}}=\frac{{dg}}{{dx}}=\frac{{dg}}{{dy}}×\frac{{dy}}{{dx}}=\frac{\mathrm{2}}{\left(\mathrm{2}−{y}\right)^{\mathrm{2}} }×\frac{{y}}{\mathrm{2}−{y}}=\frac{\mathrm{2}{y}}{\left(\mathrm{2}−{y}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$\Rightarrow{answer}\:{D} \\ $$

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