Question

URGENT!!!DERIVATA VA ROG!!!

Answer 1

Salut,

$\left(\dfrac{lnx}{\sqrt x}\right) ' =\dfrac{(lnx)'\cdot\sqrt x-lnx\cdot(\sqrt x)'}{(\sqrt x)^2}=\dfrac{\dfrac{1}x\cdot\sqrt x-lnx\cdot\dfrac{1}{2\sqrt x}}{x}=\\\\\\=\dfrac{\dfrac{1}{\sqrt x}-\dfrac{lnx}{2\sqrt x}}{x}=\dfrac{2-lnx}{2\cdot\sqrt x\cdot x}.$

Ai înțeles rezolvarea ?

Green eyes.

Answer 2

$\bigg(\dfrac{ln~x}{\sqrt{x} }\bigg)'=~?\\\\ Aplic\u{a}m~formula:~\bigg(\dfrac{f}{g}\bigg)'=\dfrac{f'\cdot g-f\cdot g'}{g^2} \\\\\bigg(\dfrac{ln~x}{\sqrt{x} } \bigg)'=\dfrac{\big(ln~x\big)'\cdot\sqrt{x} -ln~x\cdot\big(\sqrt{x} \big)'}{\big(\sqrt{x} \big)^2} =\dfrac{\dfrac{1}{x}\cdot\sqrt{x} -ln~x\cdot \dfrac{1}{2\sqrt{x} } }{x} =$$=\dfrac{\dfrac{\sqrt{x} }{x}-\dfrac{ln~x}{2\sqrt{x} } }{x}=\dfrac{\dfrac{2\sqrt{x} \cdot\sqrt{x} }{2x\sqrt{x} }-\dfrac{xln~x}{2x\sqrt{x} } }{x}=\dfrac{\dfrac{2x-xln~x}{2x\sqrt{x} } }{x} =\\\\=\dfrac{x(2-ln~x)}{2x\sqrt{x} }\cdot\dfrac{1}{x}= \dfrac{x\big(2-ln~x\big)}{2x^2\sqrt{x} }=\dfrac{2-ln~x}{2x\sqrt{x} }$

$Formule:~\big(ln~x\big)'=\dfrac{1}{x};~~~~~\big(\sqrt{x}\big)' =\dfrac{1}{2\sqrt{x} }$