Question

870) Найти производную данной функции и упростить ее:
$y=\frac{sin x-xcos x}{cos x+xsin x}$

Answer 1

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