Find the first derivative dy/dx

Have you considered "Logarithmic Differentiation"?

\(\displaystyle y = x^{ln(x)}\)

\(\displaystyle ln(y) = ln\left(x^{ln(x)}\right) = ln(x) \cdot ln(x) = \left[ln(x)\right]^{2}\)

Now what? I would consider "Implicit Differentiation" at this point.
 
I've posted this more times than I really should but I still think it's amusing!

There are two obvious mistakes we could make when differentiating \(\displaystyle f(x)^{g(x)}\):
1) Treat the "g(x)" as if it were a constant: \(\displaystyle g(x)(f(x))^{g(x)- 1}f'(x)\).
2) Treat the "f(x)" as if it were a constant: \(\displaystyle ln(f(x))f(x)^{g(x)}g'(x)\).

The "amusing" part is that the correct derivative is the sum of those two errors!
Taking the logarithm of both sides of \(\displaystyle y= f(x)^{g(x)}\) gives \(\displaystyle ln(y)= g(x)ln(f(x))\). Differentiating both sides with respect to x,
\(\displaystyle \frac{1}{y}y'= g' ln(f)+ \frac{g}{f}f'\) so that, multiplying on both sides by \(\displaystyle y= f(x)^{g(x)}\),
\(\displaystyle y'= f(x)^{g(x)}(g' ln(f)+ \frac{g}{f}f'}= ln((x)f(x)^{g(x)}g'(x)+ g(x)f(x)^{g(x)-1}f'(x)\):p
 
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