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| Derivada das Funções Trigonométricas Inversas |
| Derivative of Inverse Trigonometric Functions |
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| \(\dfrac{d}{du}\sin^{-1}(u)=\dfrac{1}{\sqrt{1-u^2}}\)
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Derivada da função arco-seno
Da figura ao lado temos
\( y=\sin\alpha \hspace{0.4cm}\Rightarrow\hspace{0.4cm} \alpha=\sin^{-1}y \hspace{0.4cm}\text{ e }\hspace{0.4cm} \cos\alpha=\sqrt{1-y^2}\)
Então,
\begin{aligned}
y =\,&\, \sin\alpha \\
\dfrac{d}{dy}y =\,&\, \dfrac{d}{dy}\sin\alpha \\
1 =\,&\, \cos\alpha\dfrac{d}{dy}\alpha = \sqrt{1-y^2}\,\dfrac{d}{dy}\sin^{-1}y\\
\dfrac{d}{dy}\sin^{-1}y =\,&\, \dfrac{1}{\sqrt{1-y^2}} \hspace{0.4cm}\text{ ou }\hspace{0.4cm} \dfrac{d}{du}\sin^{-1}(u)=\dfrac{1}{\sqrt{1-u^2}}
\end{aligned} |
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| \(\dfrac{d}{du}\cos^{-1}(u)=-\dfrac{1}{\sqrt{1-u^2}}\)
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Derivada da função arco-cosseno
Da figura ao lado temos
\( x=\cos\alpha \hspace{0.4cm}\Rightarrow\hspace{0.4cm} \alpha=\cos^{-1}x \hspace{0.4cm}\text{ e }\hspace{0.4cm} \sin\alpha=\sqrt{1-x^2}\)
Então,
\begin{aligned}
x =\,&\, \cos\alpha \\
\dfrac{d}{dx}x =\,&\, \dfrac{d}{dx}\cos\alpha \\
1 =\,&\, -\sin\alpha\dfrac{d}{dx}\alpha = -\sqrt{1-x^2}\,\dfrac{d}{dx}\cos^{-1}x\\
\dfrac{d}{dx}\cos^{-1}x =\,&\, -\dfrac{1}{\sqrt{1-x^2}} \hspace{0.4cm}\text{ ou }\hspace{0.4cm} \dfrac{d}{du}\cos^{-1}(u)=-\dfrac{1}{\sqrt{1-u^2}}
\end{aligned} |
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| \(\dfrac{d}{du}\tan^{-1}(u)=\dfrac{1}{1+u^2}\)
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Derivada da função arco-tangente
Da figura ao lado temos
\( x=\tan\alpha \hspace{0.4cm}\Rightarrow\hspace{0.4cm} \alpha=\tan^{-1}x \hspace{0.4cm}\text{ e }\hspace{0.4cm} \cos\alpha=\dfrac{1}{\sqrt{1+x^2}} \hspace{0.4cm}\Rightarrow\hspace{0.4cm} \cos^2\alpha=\dfrac{1}{1+x^2} \)
Então,
\begin{aligned}
x =\,&\, \tan\alpha \\
\dfrac{d}{dx}x =\,&\, \dfrac{d}{dx}\tan\alpha \\
1 =\,&\, \sec^2\alpha\,\dfrac{d}{dx}\alpha = (1+x^2)\,\dfrac{d}{dx}\tan^{-1}x\\
\dfrac{d}{dx}\tan^{-1}x =\,&\, \dfrac{1}{1+x^2} \hspace{0.4cm}\text{ ou }\hspace{0.4cm} \dfrac{d}{du}\tan^{-1}(u)=\dfrac{1}{1+u^2}
\end{aligned} |
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| \(\dfrac{d}{du}\cot^{-1}(u)=-\dfrac{1}{1+u^2}\)
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Derivada da função arco-cotangente
Da figura ao lado temos
\( x=\cot\alpha \hspace{0.4cm}\Rightarrow\hspace{0.4cm} \alpha=\cot^{-1}x \hspace{0.4cm}\text{ e }\hspace{0.4cm} \sin\alpha=\dfrac{1}{\sqrt{1+x^2}} \hspace{0.4cm}\Rightarrow\hspace{0.4cm} \sin^2\alpha=\dfrac{1}{1+x^2} \)
Então,
\begin{aligned}
x =\,&\, \cot\alpha \\
\dfrac{d}{dx}x =\,&\, \dfrac{d}{dx}\cot\alpha \\
1 =\,&\, -\csc^2\alpha\,\dfrac{d}{dx}\alpha = -(1+x^2)\,\dfrac{d}{dx}\cot^{-1}x\\
\dfrac{d}{dx}\cot^{-1}x =\,&\, -\dfrac{1}{1+x^2} \hspace{0.4cm}\text{ ou }\hspace{0.4cm} \dfrac{d}{du}\cot^{-1}(u)=-\dfrac{1}{1+u^2}
\end{aligned} |
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| \(\dfrac{d}{du}\sec^{-1}(u)=\dfrac{1}{|u|\sqrt{u^2-1}} \)
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Derivada da função arco-secante
Da figura ao lado temos
\( \cos\alpha=\dfrac{1}{x} \hspace{0.4cm}\Rightarrow\hspace{0.4cm} x=\dfrac{1}{\cos\alpha}\equiv\sec\alpha \)
\( x=\sec\alpha \hspace{0.4cm}\Rightarrow\hspace{0.4cm} \alpha=\sec^{-1}x \hspace{0.4cm}\text{ e }\hspace{0.4cm} \tan\alpha=\sqrt{x^2-1} \)
Então,
\begin{aligned}
x =\,&\, \sec\alpha \\
\dfrac{d}{dx}x =\,&\, \dfrac{d}{dx}\sec\alpha \\
1 =\,&\, \sec\alpha\tan\alpha\,\dfrac{d}{dx}\alpha = x\,\sqrt{x^2-1}\,\dfrac{d}{dx}\sec^{-1}x\\
\dfrac{d}{dx}\sec^{-1}x =\,&\, \dfrac{1}{x\,\sqrt{x^2-1}} \hspace{0.4cm}\text{ ou }\hspace{0.4cm} \dfrac{d}{du}\sec^{-1}(u)=\dfrac{1}{|u|\sqrt{u^2-1}}
\end{aligned} |
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| \(\dfrac{d}{du}\csc^{-1}(u)=-\dfrac{1}{|u|\sqrt{u^2-1}} \)
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Derivada da função arco-cossecante
Da figura ao lado temos
\( \sin\alpha=\dfrac{1}{y} \hspace{0.4cm}\Rightarrow\hspace{0.4cm} y=\dfrac{1}{\sin\alpha}\equiv\csc\alpha \)
\( y=\csc\alpha \hspace{0.4cm}\Rightarrow\hspace{0.4cm} \alpha=\csc^{-1}y \hspace{0.4cm}\text{ e }\hspace{0.4cm} \cot\alpha=\sqrt{y^2-1} \)
Então,
\begin{aligned}
y =\,&\, \csc\alpha \\
\dfrac{d}{dy}y =\,&\, \dfrac{d}{dy}\csc\alpha \\
1 =\,&\, -\csc\alpha\cot\alpha\,\dfrac{d}{dy}\alpha = -y\,\sqrt{y^2-1}\,\dfrac{d}{dy}\csc^{-1}y\\
\dfrac{d}{dy}\csc^{-1}y =\,&\, -\dfrac{1}{y\,\sqrt{y^2-1}} \hspace{0.4cm}\text{ ou }\hspace{0.4cm} \dfrac{d}{du}\csc^{-1}(u)=-\dfrac{1}{|u|\sqrt{u^2-1}}
\end{aligned} |
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