Помогите 465 пожалуйста

Решите задачу:

$1)\; \; y=(x^3-2x^2+3)^{17}\\\\y'=17(x^3-2x^2+3)^{16}\cdot (x^3-2x^2+3)'=17(x63-2x^2+3)^{16}\cdot (3x^2-4x)\\\\\\2)\; \; y=\sqrt{1-x^4}+\frac{1}{x^2+3}\\\\y'=\frac{1}{2\sqrt{1-x^2}\cdot}(1-x^4)'+\frac{-(x^2+3)'}{(x^2+3)^2}=\frac{1}{2\sqrt{1-x^4}}\cdot (-4x^3)-\frac{2x}{(x^2+3)^2}=\\\\=-\frac{2x^3}{\sqrt{1-x^4}}-\frac{2x}{(x^2+3)^2}\\\\\\3)\; \; y=\sqrt{4x^2+5}\; \; ,\; \; \; (\sqrt{u})'=\frac{1}{2\sqrt{u}}\cdot u'\; \; ,\; \; u=4x^2+5\\\\y'=\frac{1}{2\sqrt{4x^2+5}}\cdot (4x^2+5)'=\frac{1}{2\sqrt{4x^2+5}}\cdot 8x=\frac{4x}{\sqrt{4x^2+5}}$

$4)\; \; y=(3-x^3)^5+\sqrt{2x^2+3}\\\\y'=5(3-x^3)^4\cdot (3-x^3)'+\frac{1}{2\sqrt{2x^2+3}}\cdot (2x^2+3)'=\\\\=5(3-x^3)^4\cdot (-3x^2)+\frac{1}{2\sqrt{2x^2+3}}\cdot 4x=-15x^2\cdot (3-x^3)^4+\frac{2x}{\sqrt{2x^2+3}}$