If x^yy^x = 1, then what is \(\frac{dy}{dx}\) at (1, 1) equal to?

Formula used :

1. log mn = log m + log n 
2. log mⁿ = n log m
3. Differentiation by part: \(\frac{d}{dx}(uv) = v\frac{du}{dx}+u\frac{dv}{dx}\)
4. \(\frac{d}{dx}x^n=nx^{n-1}\)
5. \(\frac{d}{dx}log\ x= \frac{1}{x}\)

Calculation :

Given that

xyyx = 1

Taking the log on both sides

log( xyyx) = log 1

Since, log mn = log m + log n 

⇒ log x^y +  log y^x = log 1

Since log mn = n log m

⇒ y log x + x log y = 0        (∵ log 1 = 0)

Diff. with resepct to x

⇒ \(\frac{d}{dx}\)(y log x) + \(\frac{d}{dx}\) (x log y) = 0 

Using differentiation by part

⇒ \(y\frac{d}{dx}log\ x+ log \ x \frac{dy}{dx}+ x\frac{d}{dx}log\ y+ log \ y \frac{dx}{dx}=0\)

By using the formula (4) & (5)

⇒ \(\frac{y}{x}+ log \ x \frac{dy}{dx}+ \frac{x}{y}\frac{dy}{dx}+ log \ y=0\)

⇒ \(\frac{y}{x}+log\ y + \frac{dy}{dx}(\frac{x}{y}+log\ x) =0\)

⇒ \( \frac{dy}{dx}=-\frac{\frac{y}{x}+log\ y}{\frac{x}{y}+log\ x}\)

Put x = y = 1 in above differential

⇒ \(\frac{dy}{dx}=-\frac{1 + log\ 1}{1 + log \ 1}= -1\)

 \(\frac{dy}{dx}\) at (1, 1) equal to -1.