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I'm not sure, that my steps are valid, so $ln(f(x))=ln \cdot (2 \cdot ln(x))^{x/2}=x^{2} \cdot ln \cdot(ln(x))$ so I get $\frac{1}{f(x)} \cdot f'(x)$ and I have to multiply both sides with $f(x)$, am I right?

User8976
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Zauberkerl
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    I think you confused some logarithm laws. There is no way the $2\cdot$ can wander to $x^2$. – SK19 Feb 19 '18 at 13:57

2 Answers2

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HINT

Let consider

$$2 (\ln (x))^{x/2}=2e^{(x/2)\ln x}$$

then use

$$(e^{f(x)})'=e^{f(x)}f'(x)$$

user
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apply $$e^{x}$$ to both side you will have:

$$e^{f^{}(x)}=e^{2(x/2)lnx}=e^{x}.e^{lnx}=e^{x}.x$$. now get derevative from both side: $$e^{f^{}(x)}.f^{′}(x)=e^{x}x+e^{x}$$ $$f^{′}(x)=e^{-f(x)}(e^{x}x+e^{x})$$ $$f^{′}(x)=e^{-2lnx^{x/2}}(e^{x}x+e^{x})$$

shere
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