Question

derivada por regra da cadeia y = ∛1-6x ? (raiz cubica de 1-6x)

Answer 1

$\sqrt[3]{1-6x} = (1-6x)^{ \frac{1}{3} }$

y= $(1-6x)^{ \frac{1}{3}}$

u=1-6x
u'= -6

y= $(u)^{ \frac{1}{3}}$ (pela regra da potência $n . u ^{n-1}$. u')
y'= $\frac{1}{3}. u^{ \frac{1}{3}-1 }$ . u'
y'= $\frac{1}{3}. u^{ \frac{-2}{3} }$ . -6
y'= $\frac{-6}{3. \sqrt[3]{u^{2} } }$
y'= $\frac{-2}{ \sqrt[3]{u^{2} } }$
y'= $\frac{-2}{ \sqrt[3]{(1-6x)^{2} } }$