Respostas

2014-05-09T15:19:25-03:00
D )

Sem usar a regra do quociente é possível simplificar antes de derivar.

f(x)= \frac{x^3-2}{ \sqrt{x} } \\
\\f(x)= \frac{x^3-2}{x^{ \frac{1}{2} }} \\
\\f(x)=(x^3-2)x^{ -\frac{1}{2} }\\
\\f(x)=x^{ \frac{5}{2} }-2x^{- \frac{1}{2} }

f'(x)= \frac{5}{2}x^{ \frac{3}{2} }+1x^{- \frac{3}{2} }

\boxed{\boxed{f'(x)= \frac{5 \sqrt{x^3} }{2}+ \frac{1}{ \sqrt{x^3} }}}

F ) Usando a regra do quociente.

f(z)= \frac{8-z+3z^2}{2-9z}


u=8-z+3z^2~~~~~~~~~~~~u'=6z-1\\v=2-9z~~~~~~~~~~~~~~~~~~v'=-9~\\\\v^2=(2-9z)^2

f'(z)= \frac{u'.v-u.v'}{v^2}

\boxed{\boxed{f'(z)= \frac{(6z-1)(2-9z)-(8-z+3z^2)(-9)}{(2-9z)^2} }}