Question

3)Dê as derivadas das funções abaixo :
a) y= sec ( ln x)
b) y= log 3 x
c) y= sen ^3 (e ^x + x^2)
d) y= cos ^4 (e ^x + x^3)
e) y= e ^tg 2x

Answer 1

$\large\begin{array}{l} \textsf{Encontrar as derivadas das fun\c{c}\~oes:}\\\\ \textsf{a) }\mathsf{y=sec(\ell n\,x)}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}\big[sec(\ell n\,x)\big]}\qquad\textsf{(usando a Regra da Cadeia)}\\\\ \mathsf{\dfrac{dy}{dx}=sec(\ell n\,x)\,tg(\ell n\,x)\cdot \dfrac{d}{dx}(\ell n\,x)}\\\\ \mathsf{\dfrac{dy}{dx}=sec(\ell n\,x)\,tg(\ell n\,x)\cdot \dfrac{1}{x}}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=\dfrac{sec(\ell n\,x)\,tg(\ell n\,x)}{x}} \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a resposta.} \end{array}$

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$\large\begin{array}{l} \textsf{b) }\mathsf{y=\ell og(3x)}\qquad\textsf{(usando a lei de mudan\c{c}a de base)}\\\\ \mathsf{y=\dfrac{\ell n(3x)}{\ell n\,10}}\\\\ \mathsf{y=\dfrac{1}{\ell n\,10}\cdot \ell n(3x)}\\\\\\ \textsf{Derivando,}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{1}{\ell n\,10}\cdot \ell n(3x) \right )}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{\ell n\,10}\cdot \dfrac{d}{dx}\big[\ell n(3x)\big]}\qquad\textsf{(usando a Regra da Cadeia)} \end{array}$

$\large\begin{array}{l} \mathsf{\dfrac{dy}{dx}=\dfrac{1}{\ell n\,10}\cdot \dfrac{1}{3x}\cdot\dfrac{d}{dx}(3x)}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{\ell n\,10}\cdot \dfrac{1}{\diagup\!\!\!\! 3x}\cdot \diagup\!\!\!\! 3}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=\dfrac{1}{x\,\ell n\,10}} \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a resposta.} \end{array}$

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$\large\begin{array}{l} \textsf{c) }\mathsf{y=sen^3(e^x+x^2)}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}\big[sen^3(e^x+x^2)\big]}\\\\ \mathsf{\dfrac{dy}{dx}=3\,sen^2(e^x+x^2)\cdot \dfrac{d}{dx}\big[sen(e^x+x^2)]}\\\\ \mathsf{\dfrac{dy}{dx}=3\,sen^2(e^x+x^2)\cdot cos(e^x+x^2)\cdot \dfrac{d}{dx}(e^x+x^2)}\\\\ \boxed{\begin{array}{c} \mathsf{\dfrac{dy}{dx}=3\,sen^2(e^x+x^2)\cdot cos(e^x+x^2)\cdot (e^x+2x)} \end{array}}\\\\ \uparrow\\\\ \textsf{esta \'e a resposta.} \end{array}$

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$\large\begin{array}{l} \textsf{d) }\mathsf{y=cos^4(e^x+x^3)}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}\big[cos^4(e^x+x^3)\big]}\\\\ \mathsf{\dfrac{dy}{dx}=4\,cos^3(e^x+x^3)\cdot \dfrac{d}{dx}\big[cos(e^x+x^3)\big]}\\\\ \mathsf{\dfrac{dy}{dx}=4\,cos^3(e^x+x^3)\cdot \big[\!\!-sen(e^x+x^3)\big]\cdot \dfrac{d}{dx}(e^x+x^3)}\\\\ \mathsf{\dfrac{dy}{dx}=4\,cos^3(e^x+x^3)\cdot \big[\!\!-sen(e^x+x^3)\big]\cdot (e^x+3x^2)}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-4\,cos^3(e^x+x^3)\cdot sen(e^x+x^3)\cdot (e^x+3x^2)} \end{array}}\\\\ \uparrow\\\\ \textsf{esta \'e a resposta.} \end{array}$

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$\large\begin{array}{l} \textsf{e) }\mathsf{y=e^{tg\,2x}}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}\!\left(e^{tg\,2x}\right)}\\\\ \mathsf{\dfrac{dy}{dx}=e^{tg\,2x}\cdot \dfrac{d}{dx}(tg\,2x)}\\\\ \mathsf{\dfrac{dy}{dx}=e^{tg\,2x}\cdot sec^2\,2x\cdot \dfrac{d}{dx}(2x)}\\\\ \mathsf{\dfrac{dy}{dx}=e^{tg\,2x}\cdot sec^2\,2x\cdot 2}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=2e^{tg\,2x}\cdot sec^2\,2x} \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a resposta.} \end{array}$

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$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags: derivada função composta regra da cadeia cálculo diferencial

Answer 2

a) dy/dx=séc(lnx)tg(lnx)/x

b) dy/dx=1/xln10

c)dy/dx=3sen^2(e^x+x^2).cós(e^2+x^2).(e^x+2x)

d)dy/dx=-4cos^3(e^x+x^3).sen(e^x+3x^2)

e) dy/dx=2^tg^2x.sec^2.2x