Question

Seria isso, vê se consegue me ajudar

Answer 1

Primeiramente vale ressaltar as fórmulas que serão usadas para fazer a derivação:

• Regra do produto:

$\star \: f'(x) = g'(x).h(x) +g(x).h'(x) \: \star$

• Regra do quociente:

$\star \: \left[\frac{g(x)}{h(x)}\right]'=\frac{g'(x).h(x)-g(x).h'(x)}{h^2(x)} \: \star$

• Regra da cadeia:

$\star \: \frac{dy}{dx} = \frac{dy}{du}. \frac{du}{dx} \star$

Aplicando essas regras temos que:

• item a)

$a) \: f(x) = 5x {}^{7} - \frac{4}{x {}^{3} } + \sqrt[5]{x {}^{2} } + 9 \\ f '(x) = 7.5x {}^{7 - 1} - 4.( - 3).x {}^{ - 3 - 1} + \frac{2}{5} .x {}^{ \frac{2}{5} - 1} + 0 \\ f'(x) = 35x {}^{6} + \frac{12}{x {}^{4} } + \frac{2}{5x {}^{ \frac{3}{5} } }$

• item b)

$b) f(x) = \frac{2x {}^{4} }{3x {}^{2} + 5} \\ f'(x) = \frac{(2x {}^{4} ) '.(3x {}^{2} + 5) - 2x {}^{4} .(3x {}^{2} + 5)' }{(3x {}^{2} + 5) {}^{2} } \\ f'(x) = \frac{8x {}^{3}.(3x {}^{2} + 5) - 2x {}^{4} .(6x)}{(3x {}^{2} + 5) {}^{2} } \\ f'(x) = \frac{24x {}^{5} + 40x {}^{3} - 12x {}^{5} }{(3x {}^{2} + 5) {}^{2} } \\ f'(x) = \frac{12x {}^{5} + 40x {}^{3} }{(3x {}^{2} + 5) {}^{2} }$

• item c)

$c)f(x)= \cos \left(x {}^{4} - \frac{1}{x {}^{5} } \right) \\ f'(x) = \cos'\left(x {}^{4} - \frac{1}{x {}^{5} } \right).\left(x {}^{4} - \frac{1}{x {}^{5} } \right)' \\ f'(x) = - \sin\left(x {}^{4} - \frac{1}{x {}^{5} } \right). \left(4x {}^{3} + \frac{5 {}^{} }{x {}^{6} } \right )$

• Item d)

$d)f(x) = \sin(2x).e {}^{4x} \\ f'(x) = ( \sin(2x))'.e {}^{4x} + \sin(2x).(e {}^{4x} )' \\ f'(x) = \cos(2x).(2x)'.e {}^{4x} + \sin(2x).e {}^{4x} .(4x)' \\ f'(x) = 2 e {}^{4x} \cos(2x) + 4e {}^{4x} \sin(2x)$

• Item e)

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

Questão 2)

Para resolver essa questão basta derivar a função três vezes, a quantidade de tracinhos indica a quantidade de derivadas.

$2) \: f'''(x) = 3x {}^{2} .e {}^{3x} \\ f '(x) =(3x {}^{2} )'.e {}^{3x} + 3x {}^{2} .(e {}^{3x} )' \\ f'(x) = 6x.e {}^{3x} + 3x {}^{2} .e {}^{3x} .(3x)' \\ f'(x) = 6x.e {}^{3x} + 9x {}^{2} .e {}^{3x} \\ \\ f''(x) = (6x)'.e {}^{3x} + 6x.(e {}^{3x} )' + (9x {}^{2} )'.e {}^{3x} + 9x {}^{2} .(e {}^{3x}) ' \\ f''(x) = 6e {}^{3x} + 6x.e {}^{3x} .(3x)' + 18x.e {}^{3x } + 9x {}^{2} .e {}^{3x} .(3x)' \\ f''(x) = 6e {}^{3x} + 18x.e {}^{3x} + 18x.e {}^{3x} + 27x {}^{2} e {}^{3x} \\ f''(x) = 6e {}^{3x} + 36x.e{}^{3x} + 27x {}^{2} e {}^{3x} \\ \\ f'''(x) = 6.(e {}^{3x} )' + 36.(x)'.e {}^{3x} + 36x.(e {}^{3x} )' + 27.(x {}^{2} )'.e {}^{3x} + 27x {}^{2} .(e {}^{3x} )' \\ f'''(x) = 6.e {}^{3x} .(3x)' + 36e {}^{3x} + 36x.e {}^{3x} .(3x)' + 54xe {}^{3x} + 27x {}^{2} .e {}^{3x} .(3x)' \\ f'''(x) = 18e {}^{3x} + 36e {}^{3x} + 108x.e {}^{3x} + 54xe {}^{3x} + 81x {}^{2} e {}^{3x} \\ f'''(x) = 54e {}^{3x} + 162xe {}^{3x} + 81x {}^{2} e {}^{3x}$

Espero ter ajudado