Помогите решить пожалуйста производные.

Задание 1.

$\displaystyle1)\quad f(x)=\frac{1}4x^4-3x^2+5\\\\f'(x)=x^3-6x\\\\f'(x_o)=(-3)^3-6*(-3)=-27+18=\boxed{-9}\\\\\\2)\quad f(x)=\frac{x^2-1}{2x+1}\\\\f'(x)=\frac{(x^2-1)'(2x+1)-(2x+1)'(x^2-1)}{(2x+1)^2}=\\\\=\frac{2x(2x+1)-2(x^2-1)}{(2x+1)^2}=\frac{4x^2+2x-2x^2+2}{(2x+1)^2}=\frac{2x^2+2x+2}{(2x+1)^2}\\\\\\f'(x_o)=\frac{2*1+2*1+2}{(2*1+1)^2}=\frac{6}{3^2}=\frac{2*3}{3^2}=\boxed{\frac{2}3}\\\\\\3)\quad y=(2x^2+1)(4+x^3)\\\\y'=(2x^2+1)'(4+x^3)+(2x^2+1)(4+x^3)'=\\\\=4x(4+x^3)+3x^2(2x^2+1)=16x+4x^4+6x^4+3x^2=$

$\displaystyle =10x^4+3x^2+16x\\\\f'(x_o)=10+3+16=\boxed{29}\\\\\\4)\quad f(x)=2x*sinx-1\\\\f'(x)=(2x)'*sinx+2x*(sinx)'-(1)'=2sinx+2xcosx\\\\f'(x_o)=2*sin\bigg(\frac{\pi}4\bigg)+2*\frac{\pi}4*cos\bigg(\frac{\pi}4\bigg)=2*\frac{\sqrt2}2+2*\frac{\pi}4*\frac{\sqrt2}2=\\\\=\boxed{\sqrt2+\frac{\pi\sqrt2}4}$

Задание 2.
$\displaystyle 1)\quad y=4^{2x-1}\\\\y'=4^{2x-1}*ln4*(2x-1)'=\boxed{2ln4*4^{2x-1}}\\\\\\2)\quad y=cos(4x+5)\\\\y'=-sin(4x+5)*(4x+5)'=\boxed{-4sin(4x+5)}\\\\\\3)\quad y=e^{x^3}+2x\\\\y'=e^{x^3}*(x^3)'+2=\boxed{3x^2e^{x^3}+2}\\\\\\4)\quad y=\sqrt{2x^2-1}\\\\y'=\frac{1}{2\sqrt{2x^2-1}}*(2x^2-1)'=\frac{4x}{2\sqrt{2x^2-1}}=\boxed{\frac{2x}{\sqrt{2x^2-1}}}$