Cho hàm số $f\left(x\right)=x{e}^{x^2}+\ln \left(x+1\right)$ . Tính f'(0) và f"(0).

Có $s^{\text{'}}\left(t\right)={\left[15+\sqrt{2}\sin \left(4\pi t+\frac{\pi }{6}\right)\right]}^{\text{'}}$$=\sqrt{2}\cos \left(4\pi t+\frac{\pi }{6}\right)\cdot {\left(4\pi t+\frac{\pi }{6}\right)}^{\text{'}}$

$=4\pi \sqrt{2}\cos \left(4\pi t+\frac{\pi }{6}\right)$

$s^{\text{'}\text{'}}\left(t\right)={\left[4\pi \sqrt{2}\cos \left(4\pi t+\frac{\pi }{6}\right)\right]}^{\text{'}}$$=-4\pi \sqrt{2}\sin \left(4\pi t+\frac{\pi }{6}\right)\cdot {\left(4\pi t+\frac{\pi }{6}\right)}^{\text{'}}$

$=-16{\pi }^2\sqrt{2}\sin \left(4\pi t+\frac{\pi }{6}\right)$

Gia tốc của hạt tại thời điểm t = 3 giây là

a(3) =  $s^{\text{'}\text{'}}\left(3\right)=-16{\pi }^2\sqrt{2}\sin \left(12\pi +\frac{\pi }{6}\right)$ $=-8{\pi }^2\sqrt{2}\approx -111,7$m/s².

Vậy gia tốc của hạt tại thời điểm t = 3 giây khoảng – 111,7 m/s².

a) y' = (ln|2x – 1|)' = $\frac{{\left(2x-1\right)}^{\text{'}}}{2x-1}$$=\frac{2}{2x-1}$

$y^{\text{'}\text{'}}={\left(\frac{2}{2x-1}\right)}^{\text{'}}$$=2\cdot \frac{-2}{{\left(2x-1\right)}^2}$$=\frac{-4}{{\left(2x-1\right)}^2}$

Vậy $y^{\text{'}\text{'}}=\frac{-4}{{\left(2x-1\right)}^2}$

b) $y^{\text{'}}={\left[\tan \left(x+\frac{\pi }{3}\right)\right]}^{\text{'}}$$=\frac{{\left(x+\frac{\pi }{3}\right)}^{\text{'}}}{co{s}^2\left(x+\frac{\pi }{3}\right)}$$=\frac{1}{co{s}^2\left(x+\frac{\pi }{3}\right)}$$1+{\tan }^2\left(x+\frac{\pi }{3}\right)$

$y^{\text{'}\text{'}}={\left[1+{\tan }^2\left(x+\frac{\pi }{3}\right)\right]}^{\text{'}}$$=2\tan \left(x+\frac{\pi }{3}\right)\cdot {\left[\tan \left(x+\frac{\pi }{3}\right)\right]}^{\text{'}}$

$=2\tan \left(x+\frac{\pi }{3}\right)\cdot \frac{{\left(x+\frac{\pi }{3}\right)}^{\text{'}}}{co{s}^2\left(x+\frac{\pi }{3}\right)}$$=2\tan \left(x+\frac{\pi }{3}\right)\cdot \frac{1}{co{s}^2\left(x+\frac{\pi }{3}\right)}$

$=2\tan \left(x+\frac{\pi }{3}\right)\cdot \left(1+{\tan }^2\left(x+\frac{\pi }{3}\right)\right)$$=2\tan \left(x+\frac{\pi }{3}\right)+2{\tan }^3\left(x+\frac{\pi }{3}\right)$

Vậy $y^{\text{'}\text{'}}=2\tan \left(x+\frac{\pi }{3}\right)+2{\tan }^3\left(x+\frac{\pi }{3}\right)$