Question

Dê as derivadas das funções admitindo suas condições de existência :
a) y= arc senx + arc tg x + arc cos x
b) y= arc sen 2x
c) y= ln( arc sen x)
d) y= ( arc sen x) ²
e) y= (arc sen x)³

Answer 1

$\large\begin{array}{l} \textsf{Derivadas envolvendo fun\c{c}\~oes trigonom\'etricas inversas.}\\\\\\ \textsf{a) }\mathsf{y=arcsen\,x+arctg\,x+arccos\,x}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(arcsen\,x+arctg\,x+arccos\,x})\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(arcsen\,x)+\dfrac{d}{dx}(arctg\,x)+\dfrac{d}{dx}(arccos\,x})\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}+\dfrac{1}{1+x^2}+\left(-\,\dfrac{1}{\sqrt{1-x^2}}\right)}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}+\dfrac{1}{1+x^2}-\dfrac{1}{\sqrt{1-x^2}}}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=\dfrac{1}{1+x^2}} \end{array}}\qquad\textsf{com } \mathsf{-1<x<1.}\end{array}$

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$\large\begin{array}{l} \textsf{b) }\mathsf{y=arcsen\,2x}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(arcsen\,2x)}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-(2x)^2}}\cdot \dfrac{d}{dx}(2x)}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-4x^2}}\cdot 2}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-4x^2}}} \end{array}}\qquad\textsf{com }\mathsf{-\frac{1}{2}<x<\frac{1}{2}.} \end{array}$

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$\large\begin{array}{l} \textsf{c) }\mathsf{y=\ell n(arcsen\,x)}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}\left[\ell n(arcsen\,x)\right]}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{arcsen\,x}\cdot \dfrac{d}{dx}(arcsen\,x)}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{arcsen\,x}\cdot \dfrac{1}{\sqrt{1-x^2}}}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=\dfrac{1}{(arcsen\,x)\sqrt{1-x^2}}} \end{array}}\qquad\textsf{com }\mathsf{0<x<1.} \end{array}$

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$\large\begin{array}{l} \textsf{d) }\mathsf{y=(arcsen\,x)^2}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}\!\left[(arcsen\,x)^2\right]}\\\\ \mathsf{\dfrac{dy}{dx}=2(arcsen\,x)^{2-1}\cdot \dfrac{d}{dx}(arcsen\,x)}\\\\ \mathsf{\dfrac{dy}{dx}=2\,arcsen\,x\cdot \dfrac{1}{\sqrt{1-x^2}}}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=\dfrac{2\,arcsen\,x}{\sqrt{1-x^2}}} \end{array}}\qquad\textsf{com }\mathsf{-1<x<1} \end{array}$

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$\large\begin{array}{l} \textsf{e) }\mathsf{y=(arcsen\,x)^3}\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}\!\left[(arcsen\,x)^3\right]}\\\\ \mathsf{\dfrac{dy}{dx}=3(arcsen\,x)^{3-1}\cdot \dfrac{d}{dx}(arcsen\,x)}\\\\ \mathsf{\dfrac{dy}{dx}=3(arcsen\,x)^2\cdot \dfrac{1}{\sqrt{1-x^2}}}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=\dfrac{3\,arcsen^2\,x}{\sqrt{1-x^2}}} \end{array}}\qquad\textsf{com }\mathsf{-1<x<1.} \end{array}$


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


Tags: derivada função trigonométrica inversa arcsen arccos arctg arctan ln composta regra da cadeia potência cálculo diferencial