![f(x)=\frac{(lnx)^2}{2x};\ D_f:x\in\mathbb{R^+}\\\\use:\left[\frac{f(x)}{g(x)}\right]'=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\\\\f'(x)=\frac{[(lnx)^2]'\cdot2x-(lnx)^2\cdot(2x)'}{(2x)^2}=(*)\\\\\ [(lnx)^2]'=2lnx\cdot\frac{1}{x}=\frac{2lnx}{x}\\\\(2x)'=2\\\\(*)=\frac{\frac{2lnx}{x}\cdot2x-(lnx)^2\cdot2}{4x^2}=\frac{4lnx-2(lnx)^2}{4x^2}=\frac{[4lnx-2(lnx)^2]:2}{4x^2:2}=\frac{2lnx-(lnx)^2}{2x^2}](/tpl/images/0751/9332/f4cae.png)
F(x)=(lnx)²/2x
give f '(x)=??...
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