Question

derivada de y=3/senx+cosx

Answer 1

$\boxed{\frac{1}{\sin(x)} =\csc(x)}$

$\boxed{\frac{d}{dx}\csc(u)=-csc(u). \cot(u).\frac{du}{dx}}$

$\boxed{\frac{d}{dx}\cos(u)=-\sin(u).\frac{du}{dx}}$

$y=\frac{3}{\sin(x)}+\cos(x)\\ y=3\csc(x)+\cos(x)$

$\boxed{\frac{dy}{dx}=-3.csc(x). \cot(x)-\sin(x)}$