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

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
$\mathrm{ln}\left(y\right)\ast {y}^{x}\ast \frac{dy}{dx}=\mathrm{ln}\left(x\right)\ast {x}^{y}\ast \frac{dy}{dx}$
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Try first taking the logarithm of both sides.
${y}^{x}={x}^{y}$
$x\mathrm{ln}y=y\mathrm{ln}x$
Next derive both sides with respect to $x$.
$\frac{d}{dx}\left(x\mathrm{ln}y\right)=\frac{d}{dx}\left(y\mathrm{ln}x\right)$
You'll need the product rule to continue. I'll do the LHS. Think you've got the RHS? Let me know!
$\mathrm{ln}y+x\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\left(y\mathrm{ln}x\right)$
The answer ought to be equivalent to
$\frac{dy}{dx}=\frac{y}{x}\cdot \frac{y-x\mathrm{ln}y}{x-y\mathrm{ln}x}$
###### Did you like this example?
Bairaxx
Take the logs and separate.

Because $\phantom{\rule{thickmathspace}{0ex}}{y}^{x}={x}^{y}\phantom{\rule{thickmathspace}{0ex}}$, so then $\phantom{\rule{thickmathspace}{0ex}}\frac{\mathrm{ln}y}{y}=\frac{\mathrm{ln}x}{x}\phantom{\rule{thickmathspace}{0ex}}$ (unless $x=0\vee y=0$)
Now $\frac{\mathrm{d}\phantom{\rule{thickmathspace}{0ex}}}{\mathrm{d}x}\left(\frac{\mathrm{ln}x}{x}\right)=\frac{1-\mathrm{ln}x}{{x}^{2}}$
And you can take it from there.