Question

Utilize a regra de derivação para determinar:

Answer 1

Oie, Td Bom?!

$f(x) = \frac{ \sqrt[3]{x} + 5x}{ \sqrt{x} }$

$f'(x) = \frac{d}{dx} ( \frac{ \sqrt[3]{x} + 5x }{ \sqrt{x} } )$

$f'(x) = \frac{d}{dx} ( \frac{x {}^{ \frac{1}{3} } + 5x}{ \sqrt{x} } )$

$f'(x) = \frac{d}{dx} ( \frac{x {}^{ \frac{1}{3} } + 5x}{ \sqrt{x} } \: . \: \frac{ \sqrt{x} }{ \sqrt{x} } )$

$f'(x) = \frac{d}{dx} ( \frac{(x {}^{ \frac{1}{3} } + 5x) \sqrt{x} }{ \sqrt{x} \sqrt{x} } )$

$f'(x) = \frac{d}{dx} ( \frac{(x {}^{ \frac{1}{3} } + 5x) \sqrt{x} }{ x} )$

$f'(x) = \frac{d}{dx} ( \frac{x {}^{ \frac{1}{3} } \: . \: (1 + 5x {}^{ \frac{2}{3} } )x {}^{ \frac{1}{2} } }{x} )$

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

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

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

$f'(x) = \frac{d}{dx} ( \frac{1 + 5x {}^{ \frac{2}{3} } }{x {}^{ \frac{1}{6} } } )$

$f'(x) = \frac{d}{dx} ( \frac{1}{x {}^{ \frac{1}{6} } } + \frac{5x {}^{ \frac{2}{3} } }{x {}^{ \frac{1}{6} } } )$

$f'(x) = \frac{d}{dx} ( \frac{1}{x {}^{ \frac{1}{6} } } + 5 x{}^{ \frac{2}{3} - \frac{1}{6} } )$

$f'(x) = \frac{d}{dx} ( \frac{1}{x {}^{ \frac{1}{6} } } + 5x {}^{ \frac{1}{2} } )$

➭Usando a Regra da Derivação:

$\frac{d}{dx} (f + g) = \frac{d}{dx} (f) + \frac{d}{dx} (g)$

$f'(x) = \frac{d}{dx} ( \frac{1}{x {}^{ \frac{1}{6} } } ) + \frac{d}{dx} (5x {}^{ \frac{1}{2} } )$

$f'(x) = - \frac{ \frac{d}{dx}(x {}^{ \frac{1}{6} }) }{(x {}^{ \frac{1}{6} }) {}^{2} } + 5 \: . \: \frac{d}{dx} (x {}^{ \frac{1}{2} } )$

$f'(x) = - \frac{ \frac{1}{6} x {}^{ - \frac{5}{6} } }{(x {}^{ \frac{1}{6} }) {}^{2} } + 5 \: . \: \frac{1}{2} x {}^{ - \frac{1}{2} }$

$f'(x) = - \frac{ \frac{1}{6} x {}^{ - \frac{5}{6} } }{x {}^{ \frac{1}{3} } } + 5 \: . \: \frac{1}{2} \: . \: \frac{1}{x {}^{ \frac{1}{2} } }$

$f'(x) = - \frac{ \frac{1}{6} }{x {}^{ \frac{7}{6} } } + \frac{5}{2x {}^{ \frac{1}{2} } }$

$f'(x) = - \frac{1}{6x {}^{ \frac{7}{6} } } + \frac{5}{2 \sqrt{x} }$

$f'(x) = - \frac{1}{6 \sqrt[6]{x {}^{7} } } + \frac{5}{2 \sqrt{x} }$

$f'(x) = - \frac{1}{6 \: . \: |x| \: . \: \sqrt[6]{x} } + \frac{5}{2 \sqrt{x} }$

$f'(x) = - \frac{1}{6 \sqrt[6]{x} \: . \: |x| } + \frac{5}{2 \sqrt{x} }$

• Sabendo que $f'(3)$, então:

$f'(3) = - \frac{1}{6 \sqrt[6]{3} \: . \: |3| } + \frac{5}{2 \sqrt{3} }$

$f'(3) = - \frac{1}{6 \sqrt[6]{3} \: . \: 3} + \frac{5}{2 \sqrt{3} } \: . \: \frac{ \sqrt{3} }{ \sqrt{3} }$

$f'(3) = - \frac{1}{6 \sqrt[6]{3} \: . \: 3} + \frac{5 \sqrt{3} }{2 \sqrt{3} \sqrt{3} }$

$f'(3) = - \frac{1}{6 \sqrt[6]{3} \: . \: 3 } + \frac{5 \sqrt{3} }{2 \: . \: 3}$

$f'(3) = - \frac{1}{6 \sqrt[6]{3} \: . \: 3} + \frac{5 \sqrt{3} }{6}$

$f'(3) = - \frac{1}{18 \sqrt[6]{3} } + \frac{5 \sqrt{3} }{6}$

$f'(3) = - \frac{1}{18 \sqrt[6]{3} } \: . \: \frac{ \sqrt[6]{3 {}^{5} } }{ \sqrt[6]{3 {}^{5} } } + \frac{5 \sqrt{3} }{6}$

$f'(3) = - \frac{1 \sqrt[6]{3 {}^{5} } }{18 \sqrt[6]{3} \sqrt[6]{3 {}^{5} } } + \frac{5 \sqrt{3} }{6}$

$f'(3) = - \frac{ \sqrt[6]{3 {}^{5} } }{18 \sqrt[6]{3 \: . \: 3 {}^{5} } } + \frac{5 \sqrt{3} }{6}$

$f'(3) = - \frac{ \sqrt[6]{3{}^{5} } }{18 \sqrt[6]{3 {}^{6} } } + \frac{5 \sqrt{3} }{6}$

$f'(3) = - \frac{ \sqrt[6]{3 {}^{5} } }{18 \: . \: 3} + \frac{5 \sqrt{3} }{6}$

$f'(3) = - \frac{ \sqrt[6]{3 {}^{5} } }{54} + \frac{5 \sqrt{3} }{6}$

$f'(3) = - \frac{ \sqrt[6]{243} }{54} + \frac{5 \sqrt{3} }{6}$

$f'(3)≈1,39712...$

Att. Makaveli1996

Answer 2

Resposta:

1)  7 / 3

2) 2

3) 2

4) 0

5) 2

6) 12

Explicação passo-a-passo: