Question

prove que a derivada de cotgx é -cossec²x

Answer 1

$\mathrm{Regra\ do\ Produto\ \to\ \boxed{\mathrm{\dfrac{d}{dx}\bigg(\dfrac{f(x)}{g(x)}\bigg)=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}}}}$

$\mathrm{\dfrac{d}{dx}(\cot{x})=\dfrac{d}{dx}\bigg(\dfrac{\cos{x}}{\sin{x}}\bigg)=\dfrac{(\cos{x})'\sin{x}-\cos{x}(\sin{x})'}{(\sin{x})^2}=}$

$\mathrm{=\dfrac{(-\sin{x})\sin{x}-\cos{x}(\cos{x})}{\sin^2{x}}=\dfrac{-(\sin^2{x}+\cos^2{x})}{\sin^2{x}}}$

$\mathrm{\Rightarrow Lembrando\ que\ \boxed{\mathrm{\sin^2{x}+\cos^2{x}=1}}}$

$\mathrm{-\dfrac{\sin^2{x}+\cos^2{x}}{\sin^2{x}}=-\dfrac{1}{\sin^2{x}}=\boxed{\mathbf{-\csc^2{x}}}}$