A) \[1/2\]
B) \[2\]
C) \[2m/n\]
D) \[~m/2n\]
Correct Answer: B
Solution :
\[{{x}^{m}}{{y}^{n}}={{(x+y)}^{m+n}}\] Taking log on both sides, we get \[m\,\,\log \,x+n\,log\,y=(m+n)\,log(x+y)\] On differentiating w.r.t.x, we get \[\frac{m}{x}+\frac{n}{y}\,\,\frac{dy}{dx}=\frac{(m+n)}{(x+y)}\left[ 1+\frac{dy}{dx} \right]\] \[\Rightarrow \] \[\frac{dy}{dx}\left[ \frac{n}{y}-\frac{(m+n)}{(x+y)} \right]=\frac{m+n}{x+y}-\frac{m}{x}\] \[\Rightarrow \] \[\frac{dy}{dx}\left[ \frac{nx+ny-my-ny}{y(x+y)} \right]\] \[=\frac{mx+nx-mx-my}{x(x+y)}\] \[\Rightarrow \] \[\frac{dy}{dx}=\left( \frac{nx-my}{x(x+y)} \right)\left( \frac{y(x+y)}{nx-my} \right)\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{y}{x}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{\begin{smallmatrix} x=1, \\ x=2 \end{smallmatrix}}}=2\]
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