Помогите решить 2 вариант)

Решите задачу:

$1.\;f(x)=3x\sqrt[3]{x}-2x+5+\frac{2}{x^2\sqrt x}+\frac1x=\\=3x^{1+\frac13}-2x+5+2x^{-2-\frac12}+\frac1x=3x^\frac43-2x+5+2x^{-\frac52}+\frac1x\\f'(x)=\frac43\cdot3x^{\frac13}-2-\frac52\cdot2x^{-\frac72}-\frac1{x^2}=4\sqrt[3]x-2-\frac5{x^3\sqrt x}-\frac1{x^2}\\f'(1)=4-2-5-1=-4$
$2.\;f(x)=(t+1)\sqrt{t^2+1}\\f'(x)=(t+1)'\sqrt{t^2+1}+(t+1)\left(\sqrt{t^2+1}\right)'=\\=\sqrt{t^2+1}+\frac{t+1}{2\sqrt{t^2+1}}\cdot(t^2+1)'=\sqrt{t^2+1}+\frac{(t+1)\cdot2t}{2\sqrt{t^2+1}}=\\=\sqrt{t^2+1}+\frac{(t+1)\cdot t}{\sqrt{t^2+1}}\\f'(1)=\sqrt{2}+\frac{2}{\sqrt2}=\sqrt2+\frac{\sqrt{2}\cdot\sqrt2}{\sqrt2}=2\sqrt2$
$3.\;f(x)=\frac{9x}{\sqrt{x^2+1}}\\f'(x)=\frac{9\sqrt{x^2+1}-\frac{9x}{2\sqrt{x^2+1}}\cdot(x^2+1)'}{(\sqrt{x^2+1})^2}=\frac{9\sqrt{x^2+1}-\frac{9x\cdot2x}{2\sqrt{x^2+1}}}{x^2+1}=\\=\frac{9\sqrt{x^2+1}-\frac{9x^2}{\sqrt{x^2+1}}}{x^2+1}=\frac{\frac{9(x^2+1)-9x^2}{\sqrt{x^2+1}}}{x^2+1}=\frac9{(x^2+1)\sqrt{x^2+1}}\\f'(1)=\frac9{(1+1)\sqrt{1+1}}=\frac9{2\sqrt2}$