Question

Найти производные первого порядка y=sin^2x/(1+sin^2x) заранее

Answer 1

$y'=\left(\dfrac{\sin^2x}{1+\sin^2x}\right)'=\left(1-\dfrac1{1+\sin^2x}\right)'=-((1+\sin^2x)^{-1})'=\\=(1+\sin^2x)^{-2}\cdot(\sin^2x)'=(1+\sin^2x)^{-2}\cdot2\sin x\cos x=\\=\dfrac{\sin2x}{(1+\sin^2x)^2}$

Answer 2

$y=\frac{sin^2 x}{1+sin^2 x}$

$y'=(\frac{sin^2 x}{1+sin^2 x})'=(\frac{sin^2 x+1-1}{1+sin^2 x})=\\\\(\frac{sin^2 x+1}{1+sin^2 x}-\frac{1}{1+sin^2 x})'=\\\\\ (1-\frac{1}{1+sin^2 x})'=\\\\\ (1)'-(\frac{1}{1+sin^2 x})'=\\\\0-(-(\frac{1}{1+sin^2 x})^2)*(sin^2 x)=\frac{1}{(1+sin^2 x)^2}*2sin^{2-1} x *(sin x)'=\\\\\frac{2sin x}{(1+sin^2 x)^2}*cos x=\frac{sin (2x)}{(1+sin^2 x)^2}$