Question

Calcule a derivada das funções:

Answer 1

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$\large\green{\boxed{\rm~~~\red{ G)}~\gray{\left(\dfrac{sen(x)}{2x}\right)'}~\pink{=}~\blue{\dfrac{x \cdot cos(x) - sen(x)}{2x^2}}~~~}}$

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$\large\green{\boxed{\rm~~~\red{ H)}~\gray{(\sqrt[4]{x^5} - \sqrt[5]{x})'}~\pink{=}~\blue{ \dfrac{5\sqrt[4]{x}}{4} - \dfrac{1}{5\sqrt[5]{x^4}} }~~~}}$

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$\large\green{\boxed{\rm~~~\red{ I)}~\gray{(\sqrt{x} + 2x + 3x^6)'}~\pink{=}~\blue{  \dfrac{1}{2\sqrt[2]{x}} + 2 + 18x^5 }~~~}}$

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$\large\green{\boxed{\rm~~~\red{ J)}~\gray{\left(\dfrac{1}{2x + 1}\right)'}~\pink{=}~\blue{ \dfrac{-2}{4x^2 + 4x + 1} }~~~}}$

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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, User.  Vamos a mais alguns exercícios de Derivação❗ Acompanhe a resolução abaixo. ✌

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

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$\Large\gray{\boxed{\rm\blue{ \left(\dfrac{sen(x)}{2x}\right)' }}}$

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☔ Pela regra do quociente para derivação temos que

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

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$\large\sf\blue{ = \dfrac{(sen(x))' \cdot 2x - sen(x) \cdot (2x)'}{(2x)^2} }$

$\large\sf\blue{ = \dfrac{cos(x) \cdot 2x - sen(x) \cdot 2}{4x^2} }$

$\large\sf\blue{ = \dfrac{2 \cdot (x \cdot cos(x) - sen(x))}{4x^2} }$

$\large\sf\blue{ = \dfrac{x \cdot cos(x) - sen(x)}{2x^2} }$

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$\large\green{\boxed{\rm~~~\red{ G)}~\gray{\left(\dfrac{sen(x)}{2x}\right)'}~\pink{=}~\blue{\dfrac{x \cdot cos(x) - sen(x)}{2x^2}}~~~}}$ ✅

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

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$\Large\gray{\boxed{\rm\blue{ (\sqrt[4]{x^5} - \sqrt[5]{x})' }}}$

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☔ Uma das regras da Derivação nos diz que a derivada de uma subtração é equivalente à subtração das derivadas, ou seja,

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$\Large\red{\boxed{\pink{\boxed{\begin{array}{rcl}&&\\&\orange{\sf (ax - by)' = (ax)' - (by)'}&\\&&\\\end{array}}}}}$

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$\large\sf\blue{ = (\sqrt[4]{x^5})' - (\sqrt[5]{x})'}$

$\large\sf\blue{ = (x^\frac{5}{4})' - (x^\frac{1}{5})'}$

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☔ Pela regra do tombo temos que

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

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$\large\sf\blue{ = \dfrac{5}{4} \cdot x^\frac{1}{4} - \dfrac{1}{5} \cdot x^\frac{-4}{5}}$

$\large\sf\blue{ = \dfrac{5\sqrt[4]{x}}{4} - \dfrac{1}{5\sqrt[5]{x^4}}}$

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$\large\green{\boxed{\rm~~~\red{ H)}~\gray{(\sqrt[4]{x^5} - \sqrt[5]{x})'}~\pink{=}~\blue{ \dfrac{5\sqrt[4]{x}}{4} - \dfrac{1}{5\sqrt[5]{x^4}} }~~~}}$ ✅

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

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$\Large\gray{\boxed{\rm\blue{ (\sqrt{x} + 2x + 3x^6)' }}}$

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☔ Uma das regras da Derivação nos diz que a derivada de uma soma é equivalente à soma das derivadas, ou seja,

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$\Large\red{\boxed{\pink{\boxed{\begin{array}{rcl}&&\\&\orange{\sf (ax + by)' = (ax)' + (by)'}&\\&&\\\end{array}}}}}$

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$\large\sf\blue{ = (\sqrt{x})' + (2x)' + (3x^6)'}$

$\large\sf\blue{ = (x^{\frac{1}{2}})' + (2x)' + (3x^6)'}$

$\large\sf\blue{ = \dfrac{1}{2\sqrt[2]{x}} + 2x^0 + 6 \cdot 3x^5}$

$\large\sf\blue{ = \dfrac{1}{2\sqrt[2]{x}} + 2 + 18x^5}$

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$\large\green{\boxed{\rm~~~\red{ I)}~\gray{(\sqrt{x} + 2x + 3x^6)'}~\pink{=}~\blue{  \dfrac{1}{2\sqrt[2]{x}} + 2 + 18x^5 }~~~}}$ ✅

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

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$\Large\gray{\boxed{\rm\blue{ \left(\dfrac{1}{2x + 1}\right)' }}}$

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$\large\sf\blue{ = \dfrac{(\diagup\!\!\!\!{1})' \cdot (2x + 1) - 1 \cdot (2x + 1)'}{(2x + 1)^2}}$

$\large\sf\blue{ = \dfrac{-1 \cdot 2}{4x^2 + 4x + 1}}$

$\large\sf\blue{ = \dfrac{-2}{4x^2 + 4x + 1}}$

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$\large\green{\boxed{\rm~~~\red{ J)}~\gray{\left(\dfrac{1}{2x + 1}\right)'}~\pink{=}~\blue{ \dfrac{-2}{4x^2 + 4x + 1} }~~~}}$ ✅

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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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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$

Answer 2

Resposta:

Explicação passo-a-passo:

h)

f(x) = x^(5/4) - x^(1/5)

f'(x) = _5x^(5/4 -1) - _x^(1/5 - 1)_

                 4                    5

f'(x)  = _5x^(1/4)_ - _x^(-4/5)_

                 4                5  

f'(x) =  _5x^(1/4)_ - ___1__

                4            5x^(4/5)

i)

f(x) = √x + 2x + 3x^6

f'(x) = _1_ + 2 + 18x^5

        2√x

j)

f(x) = __ 1__

        2x + 1

f'(x) = _(2x + 1)(0) - 1( 2)_

                    (2x + 1)²

f'(x) =    - ___ 2____

                  (2x + 1)²