Помогите решить пожалуйста

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