Question

Derivada S(t) = √t*a*e^-t
a é constante
minha resposta deu -a/(2*√t*e^t) está correta ?

Answer 1

$\sqrt{t}*a*e^{-t} = a \sqrt{t}*e^{-t}$

$u'*v+v'*u$

$( \frac{a}{2} (t)^{ \frac{-1}{2} *e^{-t}) - (e^{-t}*a \sqrt{t})$

$ae^{-t}* \frac{1}{2 \sqrt{t}} -ae^{-t}* \sqrt{t}$

$ae^{-t}*( \frac{1}{2 \sqrt{t}}- \sqrt{t}) = ae^{-t} *( \frac{-2 \sqrt{t}^2+1}{2 \sqrt{t} }) = ae^{-t}*( \frac{-2t+1}{2 \sqrt{t} } )$

E por ai vai...

Answer 2

Oie, Td Bom?!

$S(t) = \sqrt{t} a e {}^{ - t}$

$S'(t) = \frac{d}{dt} ( \sqrt{t} ae {}^{ - t} )$

$S'(t) = \frac{d}{dt} (a \sqrt{t} e {}^{ - t} )$

$S'(t) = \frac{d}{dt} (a \sqrt{t} ) \: . \: e {}^{ - t} + a \sqrt{t} \: . \: \frac{d}{dt} (e {}^{ - t} )$

... Cálculo por partes.

I.

$= \frac{d}{dt} (a \sqrt{t} )$

$= a \: . \: \frac{d}{dt} ( \sqrt{t} )$

$= a \: . \: \frac{1}{2 \sqrt{t} }$

II.

$= \frac{d}{dt} (e {}^{ - t} )$

$= \frac{d}{dg} (e {}^{g} ) \: . \: \frac{d}{dt} ( - t)$

$= e {}^{g} \: . \: ( - 1)$

$= e {}^{ - t} \: . \: ( - 1)$

... Continuando:

$S'(t) = a \: . \: \frac{1}{2 \sqrt{t} } \: . \: e {}^{ - t} + a \sqrt{t} e {}^{ - t} \: . \: ( - 1)$

$S'(t) = \frac{ae {}^{ - t} }{2 \sqrt{t} } - a \sqrt{t} e {}^{ - t} \: . \: 1$

$S'(t) = \frac{a}{2 \sqrt{t} e {}^{t} } - a \sqrt{t} \: . \: \frac{1}{e {}^{t} }$

$S'(t) = \frac{a}{2 \sqrt{t} e {}^{t} } - \frac{a \sqrt{t} }{e {}^{t} }$

$S'(t) = \frac{a - 2at}{2 \sqrt{t} e {}^{t} }$

Att. Makaveli1996