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

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

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