Question

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

• A. $\frac{y(x+y)-x}{x(1-(x+y\left)\ln x\right)}$
• B. $\frac{y{x}^{y-1}-1}{x^y\ln x-1}$
• C. $\frac{y{x}^{y-1}+1}{x^y\ln x+1}$
• D. $\frac{y-x}{x(1-\ln x)}$

Answer 1

The correct option is A $\frac{y(x+y)-x}{x(1-(x+y\left)\ln x\right)}$

$x+y=x^y$
Taking $\ln$ both sides, we get
$\ln (x+y)=y\ln x$
Differentiating both sides with respect to $x$
$\frac{1}{x+y}(1+\frac{dy}{dx})=\ln x\frac{dy}{dx}+\frac{y}{x}$
$\Rightarrow 1+\frac{dy}{dx}=(x+y)\ln x\frac{dy}{dx}+\frac{y(x+y)}{x}$
$\Rightarrow \frac{dy}{dx}(1-(x+y\left)\ln x\right)=\frac{y(x+y)}{x}-1$
$\Rightarrow \frac{dy}{dx}=\frac{y(x+y)-x}{x(1-(x+y\left)\ln x\right)}$