Question

Derivada de ln(sec3x+tg3x)

Answer 1

$y=\mathrm{\ell n\,}(\sec 3x+\mathrm{tg\,}3x)$


Aplicando a Regra da Cadeia, temos

$\dfrac{dy}{dx}=\frac{d}{dx}\left[\mathrm{\ell n\,}(\sec 3x+\mathrm{tg\,}3x) \right ]\\ \\ \\ \dfrac{dy}{dx}=\dfrac{1}{\sec 3x+\mathrm{tg\,}3x}\cdot \frac{d}{dx}\left[\sec 3x+\mathrm{tg\,}3x \right ]\\ \\ \\ \dfrac{dy}{dx}=\dfrac{1}{\sec 3x+\mathrm{tg\,}3x}\cdot \left[\frac{d}{dx}(\sec 3x)+\frac{d}{dx}(\mathrm{tg\,}3x) \right ]\\ \\ \\ \dfrac{dy}{dx}=\dfrac{1}{\sec 3x+\mathrm{\,tg\,}3x}\cdot \left[\sec 3x\mathrm{\,tg\,}3x\cdot \frac{d}{dx}(3x)+\sec^{2} 3x\cdot \frac{d}{dx}(3x) \right ]\\ \\ \\ \dfrac{dy}{dx}=\dfrac{1}{\sec 3x+\mathrm{\,tg\,}3x}\cdot \left[\sec 3x\mathrm{\,tg\,}3x\cdot 3+\sec^{2}3x\cdot 3 \right]$


$\dfrac{dy}{dx}=\dfrac{3\sec 3x}{\sec 3x+\mathrm{\,tg\,}3x}\cdot \left[\mathrm{tg\,}3x+\sec 3x \right ]$


Simplificando o fator comum $\sec 3x+\mathrm{tg\,}3x$ no numerador e no denominador, chegamos a

$\dfrac{dy}{dx}=3\sec 3x$