EM THAM KHẢO :

13.

$y = \dfrac{x \sin x + e^x}{x + \cos x}$

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

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

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

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

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

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