Let y = \(\left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) ^{ 2 }\)=\(x+\frac { 1 }{ x } +2\)
\(\frac{d y}{d x}=1-\frac{1}{x^2}\)

Let y = x⁴ - 3 sinx + cos x
\(\frac { dy }{ dx } \) = 4x³ - 3 cos x - sin x

Let y = \(\frac { 3+2x-{ x }^{ 2 } }{ x } \)
Quotient rule \(\frac{d y}{d x}=\frac{v u^{\prime}-u v^{\prime}}{v^2}\)
\(\frac{d y}{d x}=\frac{x(2-2 x)-\left(3+2 x-x^2\right) 1}{x^2}\)
\(=\frac{2 x-2 x^2-3-2 x+x^2}{x^2}\)
\(=\frac{-3-x^2}{x^2}=\frac{-\left(3+x^2\right)}{x^2}\)

Let y = \(\sqrt { x } +\frac { 1 }{ \sqrt [ 3 ]{ x } } +{ e }^{ x }\) =​ \({ x }^{ \frac { 1 }{ 2 } }+{ x }^{ \frac { -1 }{ 3} }+{ e }^{ x }\)​​​​​​
\(\frac { dy }{ dx } =\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } -1 }-\frac { 1 }{ 3 } { x }^{ \frac { -1 }{ 3 } -1 }+{ e }^{ x }\)
=  \(\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } -1 }-\frac { 1 }{ 3 } { x }^{ \frac { -4 }{ 3 } }+{ e }^{ x }\)  \(=\frac { 1 }{ 2\sqrt { x } } -\frac { 1 }{ 3\sqrt [ 3 ]{ { x }^{ 4 } } } +{ e }^{ x }\)

Let y = \(\frac { 5 }{ { x }^{ 4 } } -\frac { 2 }{ { x }^{ 3 } } +\frac { 5 }{ x } \) = 5x^-4 - 2x^-3 + 5x^-1
dy/dx = -20x^-5 + 6x^-4 - 5x^-2
= \(-\frac { 20 }{ { x }^{ 5 } } +\frac { 6 }{ { x }^{ 4 } } -\frac { 5 }{ { x }^{ 2 } } \)

3x⁴ - 2x³ + x + 8 
dy/dx = 12x³ - 6x² + 1

Let y = \(\frac { { e }^{ x } }{ 1+{ e }^{ x } } \)
\(=\frac{e^x+e^{2 x}-e^{2 x}}{\left(1+e^x\right)^2}\)
\(\frac{d y^x}{d x}=\frac{\left(1+e^x\right) e^x-e^x \cdot e^x}{\left(1+e^x\right)^2}=\frac{e^x+e^{2 x}-e^{2 x}}{\left(1+e^2\right)^2}\)
\(=\frac{e^x}{\left(1+e^x\right)^2}\)

Let y = \(\frac { { x }^{ 2 }+x+1 }{ { x }^{ 2 }-x+1 } \)
\(\frac{d y}{d x}=\frac{\left(x^2-x+1\right)(2 x+1)-\left(x^2+x+1\right)(2 x-1)}{\left(x^2-x+1\right)^2}\)
\(2 x^3-2 x^2+2 x+x^2-x+1 \)
\(=\frac{-2 x^3-2 x^2-2 x+x^2+x+1}{\left(x^2-x+1\right)^2} \)
\(=\frac{-2 x^2+2}{\left(x^2-x+1\right)^2}=\frac{-2\left(x^2-1\right)}{\left(x^2-x+1\right)^2}\)

Let y = \(\frac { { e }^{ x } }{ 1+x } \)
\(\frac{d y}{d x}=\frac{(1+x) e^x-e^x(1)}{(1+x)^2}=\frac{e^x+x e^x-e^x}{(1+x)^2}=\frac{x e^x}{(1+x)^2}\)

y = x³e^x
\(\frac{d y}{d x}=x^x\left(e^x\right)+e^x\left(3 x^2\right)=x^2 e^x(x+3)\)