Question

Determine a derivada de F(x)= Raiz quadrada de ( sen x + 1 / 1 - sen x).

Answer 1

$f(x)=\sqrt{\dfrac{1+\sin(x)}{1-\sin(x)}}=\left(\dfrac{1+\sin(x)}{1-\sin(x)}\right)^{\frac{1}{2}}\\\\ f'(x)=\dfrac{1}{2}\left(\dfrac{1+\sin(x)}{1-\sin(x)}\right)^{-\frac{1}{2}}\cdot\left(\dfrac{1+\sin(x)}{1-\sin(x)}\right)'\\\\ f'(x)\!=\!\dfrac{1}{2}\left(\dfrac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\!\cdot\!\dfrac{(1+\sin x)'(1-\sin x)-(1+\sin x)(1-\sin x)'}{(1-\sin(x))^2}$

$f'(x)=\dfrac{1}{2}\left(\dfrac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\cdot\dfrac{\cos x(1-\sin x)-(1+\sin x)(-\cos x)}{(1-\sin(x))^2}\\\\ f'(x)=\dfrac{1}{2}\left(\dfrac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\cdot\dfrac{\cos x-\cos x\cdot\sin x+\cos x+\cos x\cdot\sin x}{(1-\sin(x))^2}\\\\ f'(x)=\dfrac{1}{2}\left(\dfrac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\cdot\dfrac{2\cos x}{(1-\sin(x))^2}\\\\ \boxed{f'(x)=\left(\dfrac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}}\cdot\dfrac{\cos x}{(1-\sin(x))^2}}$