Implicit Differentiation of $y^x=x^y$

Question

Alright so my issue is that i get stuck at this point and do not know what i should do to isolate $\frac{dy}{dx}$ since it is asking for implicit differentiation and this is what i have so far.$$\ln(y)*y^x*\frac{dy}{dx}=\ln(x)*x^y*\frac{dy}{dx}$$

Answer 1

Try first taking the logarithm of both sides. $$y^x=x^y$$ $$x \ln y=y \ln x$$

Next derive both sides with respect to $x$.

$$\frac{d}{dx}( x \ln y)=\frac{d}{dx}(y \ln x)$$

You'll need the product rule to continue. I'll do the LHS. Think you've got the RHS? Let me know!

$$\ln y + x \frac{1}{y} \frac{dy}{dx}=\frac{d}{dx}(y \ln x)$$

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The answer ought to be equivalent to

$$\frac{dy}{dx} = \frac{y}{x}\cdot\frac{y-x \ln y}{x - y \ln x}$$

Answer 2

Take the logs and separate.

Because $\;y^x=x^y\;$, so then $\;\dfrac{\ln y}{y}=\dfrac{\ln x}{x}\;$ (unless $x=0\vee y=0$)

Now $\dfrac{\mathrm d\;}{\mathrm d x}\left(\dfrac{\ln x}{x}\right) = \dfrac{1-\ln x}{x^2}$

And you can take it from there.

Answer 3

By taking the natural log of both sides, we have $x \text{ln}(y) - y \text{ln} (x)=0$

Let $$x \text{ln}(y) - y \text{ln} (x)=0\equiv F$$.

By the Implicit Function Theorem (IFT), we obtain $$\frac{dy}{dx}= - \frac{\frac{\partial F }{\partial x}}{\frac{\partial F }{\partial y}}$$

$$ \implies \frac{dy}{dx} = \frac{\frac{y}{x} - \text{ln}(y)}{\frac{x}{y}-\text{ln}(x)} = \frac{y}{x}\cdot\frac{y-x \ln (y)}{x - y \ln(x)}.$$