Question

Qual é a derivada de y=ln (tg)em x=30 graus

Answer 1

$\large\boxed{\begin{array}{l}\rm x=30^\circ=\dfrac{\pi}{6}\,rad.\\\rm y=\ell n(tg(x))\\\rm \dfrac{dy}{dx}=\dfrac{1}{tg(x)}\cdot sec^2(x)\\\\\rm\dfrac{dy}{dx}=\dfrac{\diagup\!\!\!\!\!cos(x)}{sen(x)}\cdot\dfrac{1}{\diagup\!\!\!\!\!cos^2(x)}\\\\\rm\dfrac{dy}{dx}= csc(x)\cdot cot(x)\\\\\rm\dfrac{dy}{dx}\bigg|_{x=\dfrac{\pi}{6}}=csc\bigg(\dfrac{\pi}{6}\bigg)\cdot cotg\bigg(\dfrac{\pi}{6}\bigg)\\\\\rm\dfrac{dy}{dx}\bigg|_{x=\dfrac{\pi}{6}}=2\sqrt{3}\end{array}}$