Question

determine a derivada de 3° ordem de f(x)=
$\sqrt{2x + 3}$
​

Answer 1

Resposta:

acho q é assim

Explicação:

f=x²-3

f'=2x

g=2x³

g'=6x

Answer 2

Olá,

$\tt \: f(x) = \sqrt{2x + 3} \\ \tt \: f(x) = (2x + 3 {)}^{ \frac{1}{2} }$

Derivando:

$\tt \: f'(x) = \frac{1}{2} (2x + 3 {)}^{ \frac{1}{2} - 1 } . \: 2 \\ \tt \: f'(x) = \frac{1}{ \cancel2} (2x + 3 {)}^{ - \frac{1}{2} } . \: \cancel2 \\ \tt \: f'(x) = (2x + 3 {)}^{ - \frac{1}{2} } \\ \\ \tt \: \tt \: f''(x) = - \frac{1}{2} (2x + 3 {)}^{ - \frac{1}{2} - 1 } . \: 2 \\\tt \: f''(x) = - \frac{1}{ \cancel2} (2x + 3 {)}^{ - \frac{3}{2} } . \: \cancel2 \\ \tt \: f''(x) = - (2x + 3 {)}^{ - \frac{3}{2} } \\ \\ \tt \: f'''(x) = - \left( - \frac{3}{2} \right) (2x + 3 {)}^{ - \frac{3}{2} - 1 } . \: 2 \\\tt \: f'''(x) = - \left( - \frac{3}{ \cancel2} \right) (2x + 3 {)}^{ - \frac{5}{2} } . \: \cancel2 \\ \boxed{\tt \: f'''(x) = 3 (2x + 3 {)}^{ - \frac{5}{2} }}$