Question

Найти производную данной функции и ее: y=(sin^2x)/(sinx^2)

Answer 1

$y= \frac{\sin^2x}{\sin x^2}$

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