Question

Utilize a regra de derivação para determinar:

Answer 1

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$\large\green{\boxed{\rm~~~\orange{f'(3)}~\pink{\approx}~\blue{1,4}~~~}}$

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$\bf\large\green{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}}$

$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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☺lá novamente, Airton. Vamos a mais um exercício de limite❗

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$\large\gray{\boxed{\rm\blue{ \left(\dfrac{\sqrt[3]{x} + 5x}{\sqrt{x}}\right)' }}}$

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☔ Pela regra do quociente temos que

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$\large\red{\boxed{\pink{\boxed{\rm\begin{array}{rcl}&&\\&\orange{\left(\dfrac{P}{Q}\right)' = \dfrac{P' \cdot Q - P \cdot Q'}{Q^2}}&\\&&\\\end{array}}}}}$

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ou seja,

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➡ $\sf\large\blue{ \left(\dfrac{\sqrt[3]{x} + 5x}{\sqrt{x}}\right)' = \dfrac{(\sqrt[3]{x} + 5x)' \cdot \sqrt{x} - (\sqrt[3]{x} + 5x) \cdot (\sqrt{x})'}{(\sqrt{x})^2}}$

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☔ Pela regra do tombo temos que

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$\large\red{\boxed{\pink{\boxed{\rm\begin{array}{rcl}&&\\&\orange{ (a \cdot x^n)' = a \cdot n \cdot x^{n-1} }&\\&&\\\end{array}}}}}$

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ou seja,

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➡ $\sf\large\blue{ (\sqrt[3]{x})' = (x^{\frac{1}{3}})' = \dfrac{x^{\frac{-2}{3}}}{3}}$

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➡ $\sf\large\blue{ (5x)' = 5 }$

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➡ $\sf\large\blue{ (\sqrt[2]{x})' = (x^{\frac{1}{2}})' = \dfrac{x^{\frac{-1}{2}}}{2}}$

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☔ Sabendo que a derivada da soma é igual a soma das derivadas temos que

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➡ $\sf\large\blue{ f'(x) = \dfrac{(\sqrt[3]{x})' + 5x)' \cdot \sqrt{x} - (\sqrt[3]{x} + 5x) \cdot (\sqrt{x})'}{(\sqrt{x})^2}}$

➡ $\sf\large\blue{ f'(x) = \dfrac{(\dfrac{x^{\frac{-2}{3} } } {3} + 5) \cdot x^{\frac{1}{2}} - (x^{\frac{1}{3}} + 5x) \cdot \dfrac{x^{\frac{-1}{2} } }{2}}{x}}$

➡ $\sf\large\blue{ f'(x) = \dfrac{(\dfrac{2x^{\frac{-2}{6} } } {3} + \dfrac{30}{6}) \cdot x^{\frac{1}{2}} - \dfrac{(3x^{\frac{1}{3}} + 15x)}{3} \cdot \dfrac{x^{\frac{-1}{2} } }{2}}{x}}$

➡ $\sf\large\blue{ f'(x) = \dfrac{2x^{\frac{-1}{6}} + 30x^{\frac{1}{2}} - 3x^{\frac{-1}{6}} - 15x^{\frac{1}{2}}}{6x}}$

➡ $\sf\large\blue{ f'(x) = \dfrac{15 \sqrt{x} - \dfrac{1}{\sqrt[6]{x}}}{6x}}$

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☔ Finalmente, portanto, temos que

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➡ $\sf\large\blue{ f'(3) = \dfrac{15 \sqrt{3} - \dfrac{1}{\sqrt[6]{3}}}{18}}$

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☔ Arredondando nossas raízes temos que

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➡ $\sf\large\blue{ \sqrt{3} \approx 1,73 }$

➡ $\sf\large\blue{ \sqrt[6]{3} \approx 1,2 }$

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o que nos resulta em um valor aproximado de

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➡ $\sf\large\blue{ f'(3) \approx \dfrac{15 \cdot 1,73 - \dfrac{1}{1,2}}{18}}$

➡ $\sf\large\blue{ f'(3) \approx \dfrac{25,95 - 0,83}{18}}$

➡ $\sf\large\blue{ f'(3) \approx \dfrac{25,12}{18}}$

➡ $\sf\large\blue{ f'(3) \approx \dfrac{25,12}{18}}$

➡ $\sf\large\blue{ f'(3) \approx 1,4}$

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$\large\green{\boxed{\rm~~~\orange{f'(3)}~\pink{\approx}~\blue{1,4}~~~}}$ ✅

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

☕ $\bf\large\blue{Bons\ estudos.}$

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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$\large\textit{"Absque~sudore~et~labore}$

$\large\textit{nullum~opus~perfectum~est."}$

Answer 2

                         Derivadas

Para tornar a derivada mais fácil, podemos primeiro simplificar a expressão:

$\dfrac{\sqrt[3]{x}+5x }{\sqrt{x} } \\\\\\=\dfrac{x^{1/3} +5x}{x^{1/2} } \\\\\\ \texttt{Nos separamos em fracoes homogeneas:}\\\\\dfrac{x^{1/3} }{x^{1/2} } +\dfrac{5x}{x^{1/2} } \\\\\\x^{1/3-1/2} +5x^{1-1/2} \\\\\\=\boxed{\bold{x^{-1/6} +5x^{1/2} }}$

Derivamos com base na seguinte regra:

$\boxed{\bf{(ax^{n} )'=a(n)(x^{n-1} )}}$

$(x^{-1/6} +5x^{1/2} )'\\\\\\=(x^{-1/6} )'+(5x^{1/2} )'\\\\\\=\dfrac{-1}{6} (x^{-1/6-1} )+5(\dfrac{1}{2} )(x^{1/2-1} )}\\\\\\=\boxed{\bold{\dfrac{-1}{6} (x^{-7/6} )+(\dfrac{5}{2} )(x^{-1/2}) }}}$

Substituimos x = 3:

$\dfrac{-1}{6} (3^{-7/6} )+\dfrac{5}{2} (3^{-1/2} )\\\\\\=\dfrac{-1}{6} (0,28)+2,5(0,58)\\\\\\=-0,05+1,45\\\\\\=\boxed{\bf{1,4}}$

Espero ter ajudado, muito sorte!!