Question

Matemática cálculo I Derivação

Answer 1

$\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{\frac{d}{dx}[f(x)].g(x)-f(x).\frac{d}{dx}[g(x)]}{g(x)^2}$

$\frac{d}{dx}[\frac{\sin(x)+\cos(x)}{\sin(x)-\cos(x)}]\\\\\\\frac{\frac{d}{dx}[\sin(x)+\cos(x)].(\sin(x)-\cos(x))-(\sin(x)+\cos(x)).\frac{d}{dx}[\sin(x)-\cos(x)]}{[\sin(x)-\cos(x)]^2}\\\\\\\frac{(\cos(x)-\sin(x)).(\sin(x)-\cos(x))-(\sin(x)+\cos(x)).(\cos(x)+\sin(x))]}{[\sin(x)-\cos(x)]^2}\\\\\\\frac{\cos(x).\sin(x)-\cos^2(x)-\sin^2(x)+\sin(x).\cos(x)-(\sin(x).\cos(x)+\sin^2(x)+\cos^2(x)+\sin(x).\cos(x)}{[\sin(x)-\cos(x)]^2}$

$\frac{-\cos^2(x)-\sin^2(x)-(\sin^2(x)+\cos^2(x))}{[\sin(x)-\cos(x)]^2}\\\\\\\frac{-2(\cos^2(x)+\sin^2(x))}{[\sin(x)-\cos(x)]^2}\\\\\\\frac{-2}{\sin^2(x)-2\sin(x).\cos(x)+\cos^2(x)}\\\\\\\frac{-2}{1-2\sin(x).\cos(x)}\\\\\\\frac{-2}{1-\sin(2x)}$