Tính đạo hàm của các hàm số a, y = [sin x] / [sin x – cos x] b, y = x / [căn(4 – x^2)]

a,

$y’=\dfrac{(\sin x)'(\sin x-\cos x)-\sin x(\sin x-\cos x)’}{(\sin x-\cos x)^2}$

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

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

$=\dfrac{-1}{(\sin x-\cos x)^2}$

b,

$y’=\dfrac{\sqrt{4-x^2}-x.(\sqrt{4-x^2})’}{4-x^2}$

$=\dfrac{\sqrt{4-x^2}-x.\dfrac{-2x}{2\sqrt{4-x^2}}}{4-x^2}$

$=\dfrac{\sqrt{4-x^2}+\dfrac{x^2}{\sqrt{4-x^2}}}{4-x^2}$

$=\dfrac{ 4}{(4-x^2)\sqrt{4-x^2}}$