question_answer

If \[{{x}^{m}}{{y}^{n}}={{(x+y)}^{m\,\,+\,\,n}},\] then prove that \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=0.\]

Answer:

We have, \[{{x}^{m}}{{y}^{n}}={{(x+y)}^{m+n}}\] Taking log on both sides, we get \[\log \,({{x}^{m}}{{y}^{n}})=\log \,{{(x+y)}^{m+n}}\] \[\Rightarrow \]   \[\log {{x}^{m}}+\log {{y}^{n}}=(m+n)\,\,\log \,\,(x+y)\] \[\Rightarrow \]   \[m\log x+n\log y=(m+n)\,\,\log \,\,(x+y)\] Now, differentiating both sides w.r.t. x, we get \[\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=(m+n)\cdot \frac{1}{x+y}\cdot \left( 1+\frac{dy}{dx} \right)\] \[\Rightarrow \]   \[\frac{n}{y}\frac{dy}{dx}-\left( \frac{m+n}{x+y} \right)\frac{dy}{dx}=\left( \frac{m+n}{x+y} \right)-\frac{m}{x}\] \[\Rightarrow \]   \[\left( \frac{n}{y}-\frac{m+n}{x+y} \right)\frac{dy}{dx}=\left( \frac{m+n}{x+y} \right)-\frac{m}{x}\] \[\Rightarrow \] \[\left( \frac{nx+ny-my-ny}{y\,(x+y)} \right)\frac{dy}{dx}=\frac{xm+xn-mx-my}{x\,(x+y)}\] \[\Rightarrow \]                           \[\frac{dy}{dx}=\frac{y}{x}\]                   ?(i) Again differentiating w.r.t. x, we get Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{x\frac{dy}{dx}-y}{{{x}^{2}}}=\frac{x\cdot \frac{y}{x}-y}{{{x}^{2}}}\]             [from Eq. (i)] \[=\frac{y-y}{{{x}^{2}}}=0\]        Hence proved.