Question

Найти производноя функци 1.y=x√x+x/∛x 2.y=(arcsin³x)/(ln√x²+1)+x*tgx 3.y=2t+3, y=√t³+1

Answer 1

1)
$y`(x)=(x^\frac{3}{2}+x^\frac{2}{3})`=\frac{3\sqrt{x}}{2}+\frac{2}{3x^\frac{1}{3}}$

2)
$y`(x)=\frac{\frac{(3arcsin^2(x))(ln(\sqrt{x^2+1}))}{\sqrt{1-x^2}}-\frac{arcsin^3(x)}{x^2+1}}{ln^2(\sqrt{x^2+1})}$

3)Если  y=2t+3, x=√t³+1, то:
$\left\{ {{y=2t+3}\atop{x=\sqrt{t^3+1}}}\right\\ y`(x)=\frac{dy}{dx}\\ dy=y`(t)*dt\\ dx=x`(t)*dt\\ y`(x)=\frac{y`(t)*dt}{x`(t)*dt}=\frac{y`(t)}{x`(t)}\\ y`(t)=2\\ x`(t)=\frac{3t^2}{2\sqrt{t^+1}}\\ y`(x)=\frac{2*2\sqrt{t^+1}}{3t^2}=\frac{4\sqrt{t^+1}}{3t^2}$

Если  x=2t+3, y=√t³+1, то
$\left\{ {{y=2t+3}\atop{x=\sqrt{t^3+1}}}\right\\ y`(x)=\frac{dy}{dx}\\ dy=y`(t)*dt\\ dx=x`(t)*dt\\ y`(x)=\frac{y`(t)*dt}{x`(t)*dt}=\frac{y`(t)}{x`(t)}\\ x`(t)=2\\ y`(t)=\frac{3t^2}{2\sqrt{t^+1}}\\ y`(x)=\frac{3t^2}{2*2\sqrt{t^+1}}=\frac{3t^2}{4\sqrt{t^+1}}$