EXERCISE 3 1. Determine the derivatives of the following at any point by using first principles. \( \begin{array}{lll}\text { (a) } f(x)=x & \text { (b) } f(x)=2 x & \text { (c) } g(x)=-3 x \\ \text { (d) } f(x)=4 & \text { (e) } f(x)=5 & \text { (f) } g(x)=-7 \\ \text { (g) } f(x)=x^{2} & \text { (h) } f(x)=2 x^{2} & \text { (i) } g(x)=-4 x^{2} \\ \text { (j) } f(x)=1+x^{2} & \text { (k) } f(x)=3-x^{2} & \text { (l) } g(x)=x^{3} \\ \text { (m) } f(x)=\frac{2}{x} & \text { (n) } g(x)=-\frac{3}{x} & \end{array} \) Consider \( f: x \rightarrow x^{2}-x \)

Sure! Let's determine the derivatives of each given function using first principles. The **first principles** definition of the derivative is: \[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \] We'll apply this definition to each function from **(a)** to **(n)**. --- ### **(a) \( f(x) = x \)** \[ f'(x) = \lim_{h \to 0} \frac{(x + h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \] **Derivative:** \( f'(x) = 1 \) --- ### **(b) \( f(x) = 2x \)** \[ f'(x) = \lim_{h \to 0} \frac{2(x + h) - 2x}{h} = \lim_{h \to 0} \frac{2h}{h} = 2 \] **Derivative:** \( f'(x) = 2 \) --- ### **(c) \( g(x) = -3x \)** \[ g'(x) = \lim_{h \to 0} \frac{-3(x + h) - (-3x)}{h} = \lim_{h \to 0} \frac{-3h}{h} = -3 \] **Derivative:** \( g'(x) = -3 \) --- ### **(d) \( f(x) = 4 \)** \[ f'(x) = \lim_{h \to 0} \frac{4 - 4}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \] **Derivative:** \( f'(x) = 0 \) --- ### **(e) \( f(x) = 5 \)** \[ f'(x) = \lim_{h \to 0} \frac{5 - 5}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \] **Derivative:** \( f'(x) = 0 \) --- ### **(f) \( g(x) = -7 \)** \[ g'(x) = \lim_{h \to 0} \frac{-7 - (-7)}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \] **Derivative:** \( g'(x) = 0 \) --- ### **(g) \( f(x) = x^2 \)** \[ f'(x) = \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x \] **Derivative:** \( f'(x) = 2x \) --- ### **(h) \( f(x) = 2x^2 \)** \[ f'(x) = \lim_{h \to 0} \frac{2(x + h)^2 - 2x^2}{h} = \lim_{h \to 0} \frac{4xh + 2h^2}{h} = \lim_{h \to 0} (4x + 2h) = 4x \] **Derivative:** \( f'(x) = 4x \) --- ### **(i) \( g(x) = -4x^2 \)** \[ g'(x) = \lim_{h \to 0} \frac{-4(x + h)^2 - (-4x^2)}{h} = \lim_{h \to 0} \frac{-8xh -4h^2}{h} = \lim_{h \to 0} (-8x -4h) = -8x \] **Derivative:** \( g'(x) = -8x \) --- ### **(j) \( f(x) = 1 + x^2 \)** \[ f'(x) = \lim_{h \to 0} \frac{1 + (x + h)^2 - (1 + x^2)}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x \] **Derivative:** \( f'(x) = 2x \) --- ### **(k) \( f(x) = 3 - x^2 \)** \[ f'(x) = \lim_{h \to 0} \frac{3 - (x + h)^2 - (3 - x^2)}{h} = \lim_{h \to 0} \frac{-2xh - h^2}{h} = \lim_{h \to 0} (-2x - h) = -2x \] **Derivative:** \( f'(x) = -2x \) --- ### **(l) \( g(x) = x^3 \)** \[ g'(x) = \lim_{h \to 0} \frac{(x + h)^3 - x^3}{h} = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2 \] **Derivative:** \( g'(x) = 3x^2 \) --- ### **(m) \( f(x) = \frac{2}{x} \)** First, rewrite \( f(x) \) as \( f(x) = 2x^{-1} \). \[ f'(x) = \lim_{h \to 0} \frac{\frac{2}{x + h} - \frac{2}{x}}{h} = \lim_{h \to 0} \frac{2x - 2(x + h)}{h(x)(x + h)} = \lim_{h \to 0} \frac{-2h}{h(x)(x + h)} = \lim_{h \to 0} \frac{-2}{x(x + h)} = -\frac{2}{x^2} \] **Derivative:** \( f'(x) = -\frac{2}{x^2} \) --- ### **(n) \( g(x) = -\frac{3}{x} \)** First, rewrite \( g(x) \) as \( g(x) = -3x^{-1} \). \[ g'(x) = \lim_{h \to 0} \frac{-\frac{3}{x + h} - \left(-\frac{3}{x}\right)}{h} = \lim_{h \to 0} \frac{-3x + 3(x + h)}{h(x)(x + h)} = \lim_{h \to 0} \frac{3h}{h(x)(x + h)} = \lim_{h \to 0} \frac{3}{x(x + h)} = \frac{3}{x^2} \] **Derivative:** \( g'(x) = \frac{3}{x^2} \) --- ### **Additional Consideration: \( f(x) = x^2 - x \)** Although you mentioned "Consider \( f: x \rightarrow x^2 - x \)", let's determine its derivative using first principles. \[ f'(x) = \lim_{h \to 0} \frac{(x + h)^2 - (x + h) - (x^2 - x)}{h} \] \[ = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x - h - x^2 + x}{h} \] \[ = \lim_{h \to 0} \frac{2xh + h^2 - h}{h} \] \[ = \lim_{h \to 0} (2x + h - 1) = 2x - 1 \] **Derivative:** \( f'(x) = 2x - 1 \) --- ### **Summary of Derivatives** | **Function** | **Derivative** \( f'(x) \) | |------------------------|----------------------------------| | \( f(x) = x \) | \( 1 \) | | \( f(x) = 2x \) | \( 2 \) | | \( g(x) = -3x \) | \( -3 \) | | \( f(x) = 4 \) | \( 0 \) | | \( f(x) = 5 \) | \( 0 \) | | \( g(x) = -7 \) | \( 0 \) | | \( f(x) = x^2 \) | \( 2x \) | | \( f(x) = 2x^2 \) | \( 4x \) | | \( g(x) = -4x^2 \) | \( -8x \) | | \( f(x) = 1 + x^2 \) | \( 2x \) | | \( f(x) = 3 - x^2 \) | \( -2x \) | | \( g(x) = x^3 \) | \( 3x^2 \) | | \( f(x) = \frac{2}{x} \)| \( -\frac{2}{x^2} \) | | \( g(x) = -\frac{3}{x} \)| \( \frac{3}{x^2} \) | | \( f(x) = x^2 - x \) | \( 2x - 1 \) | --- Feel free to ask if you need further explanations or assistance with any of these derivatives!