Question

If $xy\log \left(x+y\right)=1$, prove that $\frac{dy}{dx}=-\frac{y\left(x^{2}y+x+y\right)}{x\left(x{y}^2+x+y\right)}$

Answer 1

$\mathrm{We}\mathrm{have},xy\log \left(x+y\right)=1$
Differentiating it with respect to x,
$$\begin{gathered} \Rightarrow \frac{d}{dx}\left[xy\log \left(x+y\right)\right]=\frac{d}{dx}\left(1\right) \\ \Rightarrow xy\frac{d}{dx}\log \left(x+y\right)+x\log \left(x+y\right)\frac{\mathit{d}y}{\mathit{d}x}+y\log \left(x+y\right)\frac{d}{dx}\left(x\right)=0\left[\mathrm{using}\mathrm{chain}\mathrm{rule}\mathrm{and}\mathrm{product}\mathrm{rule}\right] \\ \Rightarrow xy\left(\frac{1}{x+y}\right)\frac{d}{dx}\left(x+y\right)+x\log \left(x+y\right)\frac{\mathit{d}y}{\mathit{d}x}+y\log \left(x+y\right)\left(1\right)=0 \\ \Rightarrow \left(\frac{xy}{x+y}\right)\left(1+\frac{\mathit{d}y}{\mathit{d}x}\right)+x\log \left(x+y\right)\frac{\mathit{d}y}{\mathit{d}x}+y\log \left(x+y\right)=0 \\ \Rightarrow \left(\frac{xy}{x+y}\right)\frac{\mathit{d}y}{\mathit{d}x}+\left(\frac{xy}{x+y}\right)+x\left(\frac{1}{xy}\right)\frac{\mathit{d}y}{\mathit{d}x}+y\left(\frac{1}{xy}\right)=0\left[\because xy\log \left(x+y\right)=1\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}\left[\frac{xy}{x+y}+\frac{1}{y}\right]=-\left[\frac{1}{x}+\frac{xy}{x+y}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}\left[\frac{x{y}^2+x+y}{\left(x+y\right)y}\right]=-\left[\frac{x+y+x^{2}y}{x\left(x+y\right)}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=-\left[\frac{x+y+x^{2}y}{x\left(x+y\right)}\right]\left[\frac{y\left(x+y\right)}{x{y}^2+x+y}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=-\frac{y}{x}\left(\frac{x+y+x^{2}y}{x+y+x{y}^2}\right) \end{gathered}$$
Hence proved