question_answer

If \[{{x}^{16}}{{y}^{9}}={{({{x}^{2}}+y)}^{17}},\] then prove that \[x\frac{dy}{dx}=2y.\]

Answer:

We have, \[{{x}^{16}}{{y}^{9}}={{({{x}^{2}}+y)}^{17}}\] Taking log on both sides, we get \[16\,\,\log \,\,x+9\,\,\log \,\,y=17\,\,\log \,\,({{x}^{2}}+y)\] On differentiating w.r.t. x, we get \[\frac{16}{x}+\frac{9}{y}\cdot \frac{dy}{dx}=\frac{17}{{{x}^{2}}+y}\left( 2x+\frac{dy}{dx} \right)\] \[\Rightarrow \]\[\frac{9}{y}\cdot \frac{dy}{dx}-\frac{17}{{{x}^{2}}+y}\cdot \frac{dy}{dx}=\frac{34x}{{{x}^{2}}+y}-\frac{16}{x}\] \[\Rightarrow \]\[\frac{dy}{dx}\,\,\left( \frac{9{{x}^{2}}+9y-17y}{y\,({{x}^{2}}+y)} \right)=\frac{34{{x}^{2}}-16{{x}^{2}}-16y}{x\,({{x}^{2}}+y)}\] \[\Rightarrow \]   \[\frac{dy}{dx}\,\,\left( \frac{9{{x}^{2}}-8y}{y\,({{x}^{2}}+y)} \right)=\frac{18{{x}^{2}}-16y}{x\,({{x}^{2}}+y)}\] \[\Rightarrow \]               \[\frac{dy}{dx}=\frac{2y}{x}\] \[\Rightarrow \]               \[x\frac{dy}{dx}=2y\]          Hence proved.