Question

найти производную. Нужно все три

Answer 1

$71) \ y = x^{2} + \sin x\\y' = (x^{2} + \sin x)' = (x^{2})' + (\sin x)' = 2x + \cos x$

$93) \ y = (1 - \cos 2x)\sin 2x\\y' = ((1 - \cos 2x)\sin 2x)' = (1 - \cos 2x)'\sin 2x + (1 - \cos 2x)(\sin 2x)' =\\= (1' - (\cos 2x)')\sin 2x + (1 - \cos 2x)\cos 2x \cdot (2x)' = \\= (0 + \sin 2x \cdot (2x)')\sin 2x + (1 - \cos 2x)\cos 2x \cdot 2 =\\=\sin 2x \cdot 2 \cdot \sin 2x + (1 - \cos 2x)2\cos 2x =\\= 2\sin^{2}2x + 2\cos2x - 2\cos^{2}2x = 2(\sin^{2}2x -\cos^{2}2x) + 2\cos 2x =\\= -2\cos 4x + 2\cos 2x = 2(\cos 2x - \cos 4x) = 4\sin 3x \sin x$

$112) \ y = \cos^{3}\dfrac{x}{3} \\\\y' = \bigg( \cos^{3}\dfrac{x}{3} \bigg)' = 3\cos^{2}\dfrac{x}{3} \cdot \bigg(\cos \bigg(\dfrac{x}{3} \bigg) \bigg)' = \\= 3\cos^{2}\dfrac{x}{3} \cdot \bigg(-\sin\dfrac{x}{3} \bigg) \cdot \bigg(\dfrac{x}{3} \bigg)' = -3\cos^{2}\dfrac{x}{3} \sin \dfrac{x}{3} \cdot  \dfrac{1}{3} =\\= -\cos^{2}\dfrac{x}{3} \sin \dfrac{x}{3}$

Answer 2

71)

$y = x^{2} + sin x\\\\y' = \frac{d}{dx} (x^{2} + sinx)\\\\y' = \frac{d}{dx} (x^{2}) + \frac{d}{dx} (sin x)\\\\OTBET: y' = 2x + cos x$

93)

$y = (1 - cos2x) * sin 2x\\\\y' = \frac{d}{dx} ((1 - cos 2x) * sin 2x )\\\\y' = \frac{d}{dx} (2 * \frac{1 - cos 2x}{2} * sin 2x )\\\\y' = \frac{d}{dx} (2sin x^{2} sin 2x)\\\\y' = \frac{d}{dx} (2sinx^{2}) * sin 2x + 2sin x^{2} * \frac{d}{dx} (sin 2x)\\\\y' = 2 * 2sin (x) cos (x) sin(2x) + 2sin (x)^{2} cos (2x) * 2\\\\OTBET: y' = 4sin (x) sin (3x)$

112)

$y = cos^{3} \frac{x}{3}$

$g(x) = cos (\frac{x}{3} )$

$3cos^{2} (\frac{x}{3} ) \frac{d}{dx} [cos(\frac{x}{3} )]$

$g(x) = \frac{x}{3}$

$3cos^{2} (\frac{x}{3} ) (-sin(\frac{x}{3} ) \frac{d}{dx} [\frac{x}{3} ] )$

$-cos^{2} (\frac{x}{3} ) sin (\frac{x}{3} )$

Ответ: $-cos^{2} (\frac{x}{3} ) sin (\frac{x}{3} )$