Question

Найти производные ( с решением):

Answer 1

1)

$y=\sqrt{4-x^2+\arcsin(\frac{5x}{2} )}\\y'=\frac{1}{2\sqrt{4-x^2+\arcsin(\frac{5x}{2} )}}\times\left(4-x^2+\arcsin(\frac{5x}{2} )\right)'=\frac{-2x+\frac{5}{2\sqrt{1-\frac{25x^2}{4}}}}{2\sqrt{4-x^2+\arcsin(\frac{5x}{2} )}}=\frac{-2x+\frac{5}{\sqrt{4-25x^2}}}{2\sqrt{4-x^2+\arcsin(\frac{5x}{2} )}}$

2)

$y^3-x^2=\arctan(\frac{x}{y} )\\\frac{d}{dx}(y^3-x^2)=\frac{d}{dx}(\arctan(\frac{x}{y}))\\3y^2y'-2x=\frac{\frac{d}{dx}(\frac{x}{y} ) }{1+\frac{x^2}{y^2} }\\3y^2y'-2x=\frac{\frac{y-y'x}{y^2} }{\frac{y^2+x^2}{y^2} }\\3y^2y'-2x=\frac{y-y'x}{y^2+x^2}=\frac{y}{y^2+x^2}-\frac{y'x}{y^2+x^2}\\3y^2y'+\frac{y'x}{y^2+x^2}=\frac{y}{y^2+x^2}+2x\\y'(\frac{3y^4+3x^2y^2+x}{y^2+x^2} )=\frac{y+2xy^2+2x^3}{y^2+x^2}\\y'=\frac{x\y+2xy^2+2x^3}{3y^4+3x^2y^2+x}$