Question

F(x)=2 radical din x ori (ln x-2)
F'(x)=?
Trebuie sa ajung la lnx/radical din x

Answer 1

Răspuns:

$F(x)=2\sqrt{x}\cdot(\ln x-2)\\F'(x)=2(\sqrt{x})'\cdot(\ln x-2)+2\sqrt{x}\cdot(\ln x-2)'\\F'(x)=2\cdot\dfrac{1}{2\sqrt{x}}\codt(\ln x-2)+2\cdot \dfrac{\sqrt{x}}{x}\\F'(x)=\dfrac{\ln x}{\sqrt x} -\dfrac{2}{\sqrt{x}}+\dfrac{2}{\sqrt{x}}\\F'(x)=\dfrac{\ln x}{\sqrt{x}}~Q.E.D.$

Explicație pas cu pas:

Answer 2

$\displaystyle F(x)=2\sqrt{x} (ln~x-2)\\ \\ F'(x)=((2\sqrt{x}(ln~x-2))'=(2\sqrt{x})'(ln~x-2)+2\sqrt{x} (ln~x-2)'=\\ \\ = 2(\sqrt{x} )'(ln~x-2)+2\sqrt{x} ((ln~x)'-2')=\\ \\ =\not2 \cdot \frac{1}{\not2\sqrt{x} } (ln~x-2)+2\sqrt{x}\left(\frac{1}{x} -0\right) =\frac{1}{\sqrt{x} } (ln~x-2)+2\sqrt{x} \cdot \frac{1}{x} =$

$\displaystyle =\frac{ln~x}{\sqrt{x} } -\frac{2}{\sqrt{x} } +\frac{2\sqrt{x} }{x} = \frac{ln~x}{\sqrt{x} } -\frac{2}{\sqrt{x} } +\frac{2x^{\frac{1}{2} }}{x } =\frac{ln~x}{\sqrt{x} } -\frac{2}{\sqrt{x} } +\frac{2}{x^{\frac{1}{2} }} =\\ \\ =\frac{ln~x}{\sqrt{x} } -\frac{2}{\sqrt{x} } +\frac{2}{\sqrt{x} } =\frac{ln~x}{\sqrt{x} } \\ \\ F'(x)=\frac{ln~x}{\sqrt{x} }$