Question

помогите решить пожалуйста

Answer 1

1) $f(x)= frac{9x}{ sqrt{x^2+1} } =9x*(x^2+1)^{-1/2}$
$f'(x)=9(x^2+1)^{-1/2}+9x(- frac{1}{2})(x^2+1)^{-3/2}*2x= frac{9}{sqrt{x^2+1}}- frac{9x^2}{sqrt{(x^2+1)^3} }$
$f'(2 sqrt{2} )=frac{9}{sqrt{8+1}}- frac{9*8}{sqrt{(8+1)^3} }= frac{9}{3} - frac{72}{3^3} =3- frac{72}{27}=3- frac{8}{3}= frac{1}{3}$

2) $f(x)= frac{2x+1}{sqrt{x^2+1} }=(2x+1)(x^2+1)^{-1/2}$
$f'(x)=2(x^2+1)^{-1/2}+(2x+1)(- frac{1}{2}) (x^2+1)^{-3/2}* 2x=$
$=frac{2}{sqrt{x^2+1}} - frac{x(2x+1)}{ sqrt{(x^2+1)^3} }$
$f'( sqrt{3} )=frac{2}{sqrt{3+1}} - frac{ sqrt{3} (2 sqrt{3} +1)}{ sqrt{(3+1)^3} }= frac{2}{2} - frac{2*3+ sqrt{3}}{2^3} =1- frac{6+ sqrt{3} }{8} = frac{2+ sqrt{3} }{8}$

3) $f(x)=(x^2-1) sqrt{x^2-1}=(x^2-1)^{3/2}$
$f'(x)= frac{3}{2} (x^2-1)^{1/2}*2x=3x sqrt{x^2-1}$
$f'( sqrt{2} )=3 sqrt{2} * sqrt{2-1} =3 sqrt{2}$