Question

Вычислить производные первого порядка:

Answer 1

Ответ:

а)

$y' = x'*arctgx + x * (arctgx)' - (ln\sqrt{1+x^2})' =\\=1 * arctgx + x * \frac{1}{1+x^2} -\frac{1}{\sqrt{1+x^2}}*\frac{1}{2\sqrt{1+x^2} } *2x = \\= arctg x +\frac{x}{1+x^2} -\frac{2x}{2*(1+x^2)}=arctgx + \frac{x}{1+x^2} - \frac{x}{1+x^2} = arctgx$

б)

$y' = (ln(tgx)' * sin^2x + ln(tgx) * (sin^2x)' = \\=\frac{1}{tgx}*\frac{1}{cos^2x}*sin^2x + ln(tgx) * 2sinx*cosx = \\=\frac{1}{tgx} *tg^2x + ln(tgx) * sin2x = tgx + ln(tgx)*sin2x$

в) $y' = \frac{1}{3} * \frac{(2x^2-x-1)'*\sqrt{4x+2} - (2x^2-x-1)*(\sqrt{4x+2})' }{(\sqrt{4x+2})^2} = \frac{1}{3} * \frac{(4x - 1)*\sqrt{4x+2} - (2x^2-x-1)*\frac{1}{2\sqrt{4x+2} }*4 }{4x+2} = \\= \frac{1}{3} * \frac{(4x-1)*\sqrt{4x+2} - \frac{2(2x^2-x-1)}{\sqrt{4x+2}} }{4x+2} = \frac{1}{3} * \frac{\frac{(4x-1)(\sqrt{4x+2})^2-2(2x^2-x-1)}{\sqrt{4x+2}}}{4x+2} = \\= \frac{1}{3} * \frac{\frac{(4x-1)*(4x+2)-4x^2+2x+2} {\sqrt{4x+2}}}{4x+2} = \frac{1}{3}*\frac{16x^2+8x-4x-2 - 4x^2+2x+2}{(4x+2)*\sqrt{4x+2}} =$$= \frac{1}{3} * \frac{12x^2+6x}{(4x+2)\sqrt{4x+2} } = \frac{1}{3} * \frac{6x(2x+1)}{2(2x+1)*\sqrt{4x+2} } = \frac{1}{3} * \frac{3x}{\sqrt{4x+2}} = \frac{x}{\sqrt{4x+2}}$