Question

If x^y=y^x , then show that dy/dx = y (xlogy-x)/x (ylogx-x) . ​

Answer 1

$\boxed{\huge\mathfrak{solution}}$

Given:

$\boxed{ \bold{{x}^{y} = {y}^{x} }}$

Take Log both sides ,

$\leadsto \: log( {x}^{y} ) = log( {y}^{x} ) \\ \\ \leadsto \: y \: logx = x \: logy$

Now differentiate w.r.t.X both sides ,

$\leadsto \: \frac{d}{dx} y \: logx = \frac{d}{dx} x \:logy$

Here we use the formula:

$\boxed{ \red{\bold{\frac{d}{dx} u.v = u \: \frac{d}{dx} v + v \: \frac{d}{dx} u}}}$

$\leadsto \: y \: \frac{d}{dx} logx + logx \: \frac{dy}{dx} = x \: \frac{d}{dx} logy + logy \: \frac{d}{dx} x \\ \\ \leadsto \: y \times \frac{1}{x} + logx \: \frac{dy}{dx} = x \times \frac{1}{y} \frac{dy}{dx} + logy \times 1 \\ \\ \leadsto \: \frac{y}{x} + logx \: \frac{dy}{dx} = \frac{x}{y} \frac{dy}{dx} + logy \\ \\ \leadsto \: logx \frac{dy}{dx} - \frac{x}{y} \frac{dy}{dx} = logy - \frac{y}{x} \\ \\ \leadsto \: \frac{dy}{dx} (logx - \frac{x}{y} ) = logy - \frac{y}{x} \\ \\ \leadsto \frac{dy}{dx} ( \frac{y \: logx - x}{y} )= \frac{x \: logy - y}{x} \\ \\ \leadsto \: \frac{dy}{dx} = \frac{ \frac{x \:logy - y}{x} }{ \frac{y \: logx - x}{y} } \\ \\ \leadsto \: \frac{dy}{dx} = \frac{y}{x} ( \frac{x \: logy - y}{y \: logx - x} )$

$\huge\bold{proved}$

Answer 2

Step-by-step explanation:

Both side take log

Ylogx=xlogy

Differentiate w.r. to x

Yd(logx)/dx+logx d(y)/dx=xd(logy)dy*dy/dx+logyd(x)/dx

Y*1/x+logx dy/dx =x*1/y*dy/dx+logy

Y/x -logy=dy/dx(x/y-log x)

Y-xlogy/x=dy/dx(x-ylogx)/y

Y/x(y-xlogy)/(x-ylogx)=dy/dx proved