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Usamos a regra da cadeia para desvendar as derivadas das funções trigonométricas.
Até este ponto do curso, estivemos ignorando uma grande classe de funções: as funções trigonométricas além de \(\sin (x)\). Sabemos que \[ \ddx \sin (x) = \cos (x). \]
Munidos desse fato, vamos descobrir as derivadas de todas as funções trigonométricas usuais.
Calcule: \[ \eval {\ddx \cos \left ( \frac {x^3}{2} \right )}_{x=\sqrt [3]{\pi }} \]
Agora que conhecemos a derivada do cosseno, podemos combiná-la com a
regra da cadeia, de modo que \[ \ddx \cos \left ( \frac {x^3}{2} \right ) = \class{xm-given}{\frac {3 x^2}{2}} \left (- \sin \left ( \frac {x^3}{2} \right ) \right ) \] e, portanto,
\begin{align*} \eval {\ddx \cos \left ( \frac {x^3}{2} \right )}_{x=\sqrt [3]{\pi }} &= \eval {\left ( \frac {3}{2} x^2 \left (- \sin \left ( \frac {x^3}{2} \right ) \right ) \right )}_{x=\sqrt [3]{\pi }} \\ &= - \frac {3}{2}(\sqrt [3]{\pi })^2 \sin \left ( \frac {\pi }{2} \right ) \\ &= -\frac {3}{2} \pi ^{\frac {2}{3}} \cdot \class{xm-given}{1} \\ &=\class{xm-given}{\frac {-3 \pi ^{\frac {2}{3}}}{2}}. \end{align*}
Em seguida, temos:
A derivada da tangente \[ \ddx \tan (x) = \sec ^2(x). \]
Vamos reescrever \(\tan (x)\) como \(\frac {\sin (x)}{\cos (x)}\) e usar a regra do quociente. Escreva comigo: \begin{align*} \ddx \tan (x) &= \ddx \frac {\sin (x)}{\cos (x)}\\ &=\frac {\cos ^2(x) + \class{xm-given}{\sin ^2(x)}}{\cos ^2(x)}\\ &=\frac {\class{xm-given}{1}}{\cos ^2(x)}\\ &=\sec ^2(x). \end{align*}
Calcule: \[ \ddx \left ( \frac {5x \tan (x)}{x^2 - 3} \right ) \]
Aplicando a regra do quociente, a regra do produto e a derivada da
tangente: \begin{align*} \ddx &\left ( \frac {5x \tan (x)}{x^2 - 3} \right ) \\ &= \frac {(x^2 - 3) \cdot \ddx (\class{xm-given}{5x \tan (x)}) - 5x \tan (x) \cdot \ddx (\class{xm-given}{x^2 - 3})}{(x^2 - 3)^2} \\ &= \frac {(x^2 - 3)(5 \tan (x) + 5x \class{xm-given}{\sec ^2(x)}) - 5x \tan (x) \cdot 2x}{(x^2 - 3)^2} \\ &= \frac {5(x^2-3)(\tan (x)+x \sec ^2(x)) - 10x^2 \tan (x)}{(x^2-3)^2} \end{align*}
Por fim, temos:
A derivada da secante \[ \ddx \sec (x) = \sec (x)\tan (x). \]
Vamos reescrever \(\sec (x)\) como \((\cos (x))^{-1}\) e usar a regra da potência e a regra da cadeia. Escreva
\begin{align*} \ddx \sec (x) &= \ddx (\cos (x))^{-1}\\ &=-1(\cos (x))^{-2}(\class{xm-given}{-\sin (x)}) \\ &= \frac {\sin (x)}{\cos ^2(x)} \\ &= \frac {1}{\cos (x)} \cdot \frac {\sin (x)}{\cos (x)} \\ &= \sec (x)\tan (x). \end{align*}
As derivadas da cotangente e da cossecante são semelhantes e ficam como exercícios. Juntando tudo isso, temos:
Derivadas das funções trigonométricas
- \(\ddx \sin (x) = \cos (x)\).
- \(\ddx \cos (x) = -\sin (x)\).
- \(\ddx \tan (x) = \sec ^2(x)\).
- \(\ddx \sec (x) = \sec (x)\tan (x)\).
- \(\ddx \csc (x) = -\csc (x)\cot (x)\).
- \(\ddx \cot (x) = -\csc ^2(x)\).
Calcule: \[ \eval {\ddx ( \csc (x) \cot (x) )}_{x=\frac {\pi }{3}} \]
Aplicando a regra do produto e os fatos acima, sabemos que \[ \ddx ( \csc (x) \cot (x) ) = - \csc ^3(x) - \cot ^2(x)\class{xm-given}{\csc (x)} \] e, portanto, \[ \eval {\ddx ( \csc (x) \cot (x) )}_{x=\frac {\pi }{3}} \]
\begin{align*} &= \eval { - \csc ^3(x) - \cot ^2(x) \class{xm-given}{\csc (x)}}_{x=\frac {\pi }{3}} \\ &= - \frac {8}{3 \sqrt {3}} - \frac {1}{3}\cdot \class{xm-given}{2/\sqrt {3}} \end{align*}
Ao trabalhar com derivadas de funções trigonométricas, sugerimos que você use
radianos como medida de ângulo. Por exemplo, embora \[ \sin \left ((90^\circ \right )^2) = \sin \left (\left (\frac {\pi }{2}\right )^2\right ), \] é preciso ter cuidado com
as derivadas, pois \[ \eval {\ddx \sin \left (x^2\right )}_{x=90^\circ } \ne \underbrace {2\cdot 90\cdot \cos (90^2)}_{\text {incorreto}} \] Alternativamente, poderíamos interpretar \(x^\circ \) como significando \(\frac {x\cdot \pi }{180}\), já
que então \(90^\circ = \frac {90\cdot \pi }{180} = \frac {\pi }{2}\). Nesse caso, \[ 2\cdot 90^\circ \cdot \cos ((90^\circ )^2) = 2\cdot \frac {\pi }{2}\cdot \cos \left (\left (\frac {\pi }{2}\right )^2\right ). \]