Q12    Find dy/dx of the functions given in Exercises 12 to 15

           x ^ y + y ^ x = 1.

Answers (1)

Given function is
f(x)=x ^ y + y ^ x = 1
Now, take t = x^y
take log on both sides
\log t = y\log x
Now, differentiate w.r.t  x
we get,
\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})\\ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})
Similarly, take  k = y^x
Now, take log on both sides
\log k = x\log y
Now, differentiate w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\ \frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})
Now,
f^{'}(x)= \frac{dt}{dx}+\frac{dk}{dx}= 0

( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) + (y^x)(\log y+\frac{x}{y}\frac{dy}{dx}) = 0\\ \frac{dy}{dx}(x^y(\log x)+xy^{x-1}) = -(yx^{y-1}+y^x(\log y))\\ \frac{dy}{dx}= \frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}

Therefore, the answer is  \frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})} 

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