Question

alquem me ajuda nessa derivada preciso da resolucao ok ?

Answer 1

Boa noite.

$\frac{d}{du} (u-\sqrt{u} )(u+\sqrt{u} ) =$

Regra do produto: (ab)' = ba' +ab'

Façamos a = u-√u

$= (u+\sqrt{u} )(\frac{d}{du} (u-\sqrt{u} ))+(u-\sqrt{u} )(\frac{d}{du} (u+\sqrt{u} ))$

$=(u+\sqrt{u} )(\frac{d}{du} u-\frac{d}{du} \sqrt{u} )+(u-\sqrt{u} )(\frac{d}{du} (u+\frac{d}{du} \sqrt{u} ))$

$=(u+\sqrt{u} )(1-\frac{1}{2\sqrt{u} } )+(u-\sqrt{u} )(1+\frac{1}{2\sqrt{u} } )$

$= u-\frac{u}{2\sqrt{u} } +\sqrt{u} -\frac{\sqrt{u}}{2\sqrt{u} } +u+\frac{u}{2\sqrt{u} } -\sqrt{u} -\frac{\sqrt{u} }{2\sqrt{u} }$

$=2u-2(\frac{\sqrt{u} }{2\sqrt{u} } )$

$=2u-1$

Cálculos:

$\frac{d}{du} \sqrt{u} = \frac{d}{du} u^{\frac{1}{2} } =\frac{1}{2} u^{\frac{1}{2} -1} =\frac{1}{2} u^{-\frac{1}{2} } =\frac{1}{2\sqrt{u} }$

$\frac{d}{du} u=1$

Bons estudos. ;)