Question

If $x^y=x{e}^{x-y+6}$, then $\frac{dy}{dx}=$

• A. $\frac{x-y}{x\left(1+{\log }_{e}x\right)}$
• B. $\frac{1+x-y}{x\left(1+{\log }_{e}x\right)}$
• C. $\frac{y}{x\left(1+{\log }_{e}x\right)}$
• D. None of these

Answer 1

The correct option is B

$\frac{1+x-y}{x\left(1+{\log }_{e}x\right)}$

Explanation to the correct option.

$\begin{array}{rcl}{x}^y& =& x{e}^{x-y+6}\\ & \Rightarrow & \frac{x^y}{x}=e^{x-y+6}\\ & \Rightarrow & x^{y-1}=e^{x-y+6}\end{array}$

By taking log on both sides we get,

$$\begin{gathered} \left(y-1\right)\log x=\left(x-y+6\right)\log e \\ \Rightarrow \left(y-1\right)\log x=\left(x-y+6\right) \end{gathered}$$

By differentiating w.r.t $x$, we get

$\begin{array}{rcl}\left(y-1\right)\frac{1}{x}+\log x\frac{dy}{dx}& =& 1-\frac{dy}{dx}\\ \Rightarrow \log x\frac{dy}{dx}+\frac{dy}{dx}& =& 1-\left(\frac{y-1}{x}\right)\\ \Rightarrow \left(1+\log x\right)\frac{dy}{dx}& =& \frac{x-y+1}{x}\\ \Rightarrow \frac{dy}{dx}& =& \frac{x-y+1}{x\left(1+{\log }_{e}x\right)}\end{array}$

Hence, option B is correct