Question

Помогите найти производную функцию:
1) y=arcsin ((2x^2)/(1+x^4))
2) y=корень из 1+lnx
3) y=(1+корень из x)/(1-корень из x)

Answer 1

$1); ; y=arcsin frac{2x^2}{1+x^4}\\y'= frac{1}{sqrt{1-( frac{2x^2}{1+x^4} })^2} cdot frac{4x(1+x^4)-2x^2cdot 4x^3}{(1+x^4)^2}= frac{1+x^4}{sqrt{1+2x^4+x^8-4x^4}}cdot frac{4x+4x^5-8x^5}{(1+x^4)^2}$

$= frac{1+x^4}{sqrt{x^8-2x^4+1}} cdot frac{4x-4x^5}{(1+x^4)^2} = frac{1}{sqrt{(1-x^4)^2}} cdot frac{4x(1-x^4)}{1+x^4} = frac{4x(1-x^4)}{(1-x^4)(1+x^4)} =\\=frac{4x}{1+x^4}$

$2); ; y=sqrt{1+lnx}\\y'= frac{1}{2sqrt{1+lnx}} cdot frac{1}{x} \\3); ; y= frac{1+sqrt{x}}{1-sqrt{x}} \\y'= frac{frac{1}{2sqrt{x}}cdot (1-sqrt{x})-(1+sqrt{x})cdot (-frac{1}{2sqrt{x}})}{(1-sqrt{x})^2} = frac{1-sqrt{x}+1+sqrt{x}}{2sqrt{x}(1-sqrt{x})^2} = frac{1}{sqrt{x}(1-sqrt{x})^2}$