a) \(∆y = f(x+∆x) - f(x) = 2(x+∆x) - 5 - (2x - 5) = 2x+2∆x-5-2x+5=2∆x\)

\({{\Delta y} \over {\Delta x}} = {{2\Delta x} \over {\Delta x}} = 2\).

b) \(\Delta y = f(\Delta x + x) - f(x)\)

\(= {(x + \Delta x)^2} - 1 - ({x^2} - 1)\)

\(=x^2+2x.∆x+(∆x)^2-1-x^2+1\)

\(= 2x.\Delta x + {(\Delta x)^2}\)

\(= \Delta x(2x + \Delta x)\)

\({{\Delta y} \over {\Delta x}} = {{\Delta x\left( {2x + \Delta x} \right)} \over {\Delta x}} = 2x+\Delta x\)

c) \(∆y = f(x+∆x) - f(x) = 2(x +  ∆x)^3- 2x^3\)

\(=2\left [ x^3+3x^2.∆x+3.x(∆x)^2+(∆x)^3 \right ]-2x^3\)

\(= 2x^3+6x^2.∆x+6.x(∆x)^2+2(∆x)^3 -2x^3\)

\(=6{x^2}\Delta x + 6x{(\Delta x)^2} + 2{(\Delta x)^3}\)

\(= 2\Delta x.\left [ 3{x^2} + 3x\Delta x + {(\Delta x)^2} \right ]\)

\(\frac{\Delta y}{\Delta x} = \frac{2\Delta x\left [ 3x^2+3x\Delta x+(\Delta x)^2 \right ]}{\Delta x}= 6x^2+ 6x∆x + 2(∆x)^2\)

d) \(∆y = f(x+∆x) - f(x) =-{1 \over x} + {1 \over {x +\Delta x}} = {{-x - \Delta x + x} \over {x\left( {x + \Delta x} \right)}} =  - {{\Delta x} \over {x\left( {x + \Delta x} \right)}}\)

\({{\Delta y} \over {\Delta x}} = \frac{- {{\Delta x} \over {x\left( {x + \Delta x} \right)}}}{\Delta x}=-{1 \over {\left( {x + \Delta x} \right)x}}\)