Exercícios Resolvidos-2023 1

1o) seja \quad g(x) = \frac{1 - \cos(\frac{x}{3})}{\sen(\frac{x}{3})} \quad \quad Calcule \quad g'(\pi) 2 \circ) seja \quad g(\theta) = ln(1 + tg(\theta)) - sec^{2}(90) \quad \quad Calcule \quad g'(\frac{\pi}{4}) 3o) Calcule \quad g'(x), \quad sendo \quad \quad g(x) = \frac{sec^{2}(x)}{x\sqrt{x}} 1) \quad g(x) = \frac{1 - \cos(\frac{x}{3})}{\sen(\frac{x}{3})} *Fórmula da divisão: \frac{g}{Q(X)} = P(X) \Rightarrow g'(x) = P'(x)Q(x) - P(x)Q'(x) \begin{matrix} (Q(X) & [RQ(x)]^{2} \end{matrix} => g'(x) = \frac{\frac{1}{3} \sen x \frac{1}{3} \sen x - \frac{3}{3} \cos \frac{x}{3} (1 - \cos x)}{\sen \frac{x}{3}} \quad \quad \quad \quad (\text{SE ATENTA A REGRS DA CADEIA}) => g'(\pi) = \frac{3}{4} - \frac{1}{2} \cdot \frac{3}{4} (1 - \frac{3}{2}) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{9} \cdot \frac{2}{3} = \frac{3}{2}\underline{\frac{9}{4}} 2) \quad g(\theta) = ln(1 + tg \theta) - sec^{2}(90) => g'(\theta) = \frac{1}{7 + tg \theta} \cdot sec^{2}( \theta) - \sec 90 \cdot \text{ln} \sec 90 \cdot 9 \cdot \sqrt{3} => g' \left(\frac{\pi}{4}\right) = \frac{1}{2} \cdot 2 - 2 \sqrt{1} \cdot 9 \cdot \sqrt{3} \cdot 9 = 1 \cdot 36 = -36 \, \text{SE ATENTA A REGRS DA CADEIA}! 3) \quad g(x) = \frac{\sec^{2} x}{\frac{x}\sqrt{x}} \text{Aplicando fórmula da divisão:} => g'(x) = 2 \sec + 2 \cdot \sec x \sec x \cdot \cdot \sqrt{x} - \frac{3}{2} x^{\frac{3}{2}} \sec^{2} x \begin{matrix} \frac{\underline{x \underline{3}}}{\underline{x}^{3}} \end{matrix} = \frac{2 \sec + 9 x \cdot \sqrt{x} - \frac{3}{2} x^{\frac{3}{2}} \sec^{2}}{x^{3}}