Use the product rule:

\(f'(x)=(e^x)'(x^2+2x+4)+e^x(x^2+2x+4)'\).

Since \((e^x)'=e^x\) and \((x^2+2x+4)'=2x+2\), we get:

\(f'(x)=e^x(x^2+2x+4)+e^x(2x+2)=e^x(x^2+4x+6)\).

Substitute \(x=0\):

\(f'(0)=e^0(0^2+4\cdot0+6)=1\cdot6=6\).

Use the chain rule:

\(f'(x)=4(x^2-3x+3)^3\cdot(2x-3)\).

Substitute \(x=1\):

\(f'(1)=4(1-3+3)^3\cdot(2-3)=4\cdot1^3\cdot(-1)=-4\).

Use the quotient rule:

\(f'(x)=\dfrac{(x-3)\cdot1-x\cdot1}{(x-3)^2}=\dfrac{x-3-x}{(x-3)^2}=\dfrac{-3}{(x-3)^2}\).

Substitute \(x=4\):

\(f'(4)=\dfrac{-3}{(4-3)^2}=-3\).

Use the chain rule:

\(f'(x)=2e^{x^2-2x}\cdot(2x-2)\).

Substitute \(x=2\):

\(f'(2)=2e^{4-4}\cdot(4-2)=2e^0\cdot2=2\cdot1\cdot2=4\).

Use the product rule:

\(f'(x)=e^x(3x-5)+e^x\cdot3=e^x(3x-2)\).

Substitute \(x=0\):

\(f'(0)=e^0(3\cdot0-2)=1\cdot(-2)=-2\).

Use the chain rule:

\(f'(x)=3(x^2+2x+2)^2\cdot(2x+2)\).

Substitute \(x=-1\):

\(f'(-1)=3((-1)^2+2(-1)+2)^2\cdot(2(-1)+2)=3\cdot1^2\cdot0=0\).

Differentiate term by term:

\((5x)'=5\),

\(\left(\dfrac{4}{x-4}\right)'=4(x-4)^{-1}{}'=-4(x-4)^{-2}=\dfrac{-4}{(x-4)^2}\).

So

\(f'(x)=5-\dfrac{4}{(x-4)^2}\).

Substitute \(x=5\):

\(f'(5)=5-\dfrac{4}{1^2}=1\).

Use the chain rule:

\(f'(x)=e^{x^2-2x-8}\cdot(2x-2)\).

Substitute \(x=4\):

\(f'(4)=e^{16-8-8}\cdot(8-2)=e^0\cdot6=6\).

Using the product rule:

\(f'(x)=e^x(x^2-2x+5)+e^x(2x-2)=e^x(x^2+3)\).

Substitute \(x=0\):

\(f'(0)=e^0(0^2+3)=3\).

Use the chain rule:

\(f'(x)=4(x^2+x+1)^3\cdot(2x+1)\).

Substitute \(x=-1\):

\(f'(-1)=4(1-1+1)^3\cdot(-2+1)=4\cdot1\cdot(-1)=-4\).

Use the quotient rule:

\(f'(x)=\dfrac{(x-2)\cdot1-(x-4)\cdot1}{(x-2)^2}=\dfrac{x-2-x+4}{(x-2)^2}=\dfrac{2}{(x-2)^2}\).

Substitute \(x=3\):

\(f'(3)=\dfrac{2}{(3-2)^2}=2\).

Use the chain rule:

\(f'(x)=3e^{x^2-25}\cdot2x=6xe^{x^2-25}\).

Substitute \(x=5\):

\(f'(5)=6\cdot5\cdot e^{25-25}=30e^0=30\).

Using the product rule:

\(f'(x)=e^x(x-3)+e^x=e^x(x-2)\).

Substitute \(x=0\):

\(f'(0)=e^0(0-2)=-2\).

Use the chain rule:

\(f'(x)=3(x^2+x-3)^2\cdot(2x+1)\).

Substitute \(x=1\):

\(f'(1)=3(1+1-3)^2\cdot(2+1)=3\cdot(-1)^2\cdot3=9\).

Differentiate term by term:

\(\left(\dfrac{6}{x-6}\right)'=6(x-6)^{-1}{}'=-6(x-6)^{-2}=\dfrac{-6}{(x-6)^2}\),

\((-x)'=-1\), \((2)'=0\).

Therefore,

\(f'(x)=\dfrac{-6}{(x-6)^2}-1\).

Substitute \(x=7\):

\(f'(7)=\dfrac{-6}{1^2}-1=-7\).

Use the chain rule:

\(f'(x)=e^{x^2-2x-3}\cdot(2x-2)\).

Substitute \(x=3\):

\(f'(3)=e^{9-6-3}\cdot(6-2)=e^0\cdot4=4\).