The Derivative as a Function: Concepts, Definitions, and Differentiability

  • ## Interpretations of the Derivative at a Point f′(a)f'(a)
    • f′(a)f'(a) is the slope of the tangent line to the graph of ff at x=ax = a.
    • f′(a)f'(a) is the instantaneous rate of change of the function ff at x=ax = a.
    • If f(t)f(t) is a position function, then f′(a)f'(a) is the instantaneous velocity at time t=at = a.
  • ## Definition of the Derivative as a Function
    • If ff is a function, the derivative of ff, denoted f′f', is defined as:
      f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
    • This definition provides a function f′(x)f'(x) that gives the derivative for all values of xx where it is defined.
  • ## Example: Finding the Derivative and Tangent Line Equation
    • Problem: Find the derivative of f(x)=32x−1f(x) = \frac{3}{2x - 1} and the equation of the tangent line at x=1x = 1.
    • Step 1: Find f′(x)f'(x) using the limit definition.
    • f(x+h)=32(x+h)−1=32x+2h−1f(x + h) = \frac{3}{2(x + h) - 1} = \frac{3}{2x + 2h - 1}
    • f(x+h)−f(x)=32x+2h−1−32x−1=3(2x−1)−3(2x+2h−1)(2x+2h−1)(2x−1)f(x + h) - f(x) = \frac{3}{2x + 2h - 1} - \frac{3}{2x - 1} = \frac{3(2x - 1) - 3(2x + 2h - 1)}{(2x + 2h - 1)(2x - 1)}
    • =6x−3−(6x+6h−3)(2x+2h−1)(2x−1)=−6h(2x+2h−1)(2x−1)= \frac{6x - 3 - (6x + 6h - 3)}{(2x + 2h - 1)(2x - 1)} = \frac{-6h}{(2x + 2h - 1)(2x - 1)}
    • racf(x+h)−f(x)h=−6hh(2x+2h−1)(2x−1)=−6(2x+2h−1)(2x−1)rac{f(x + h) - f(x)}{h} = \frac{-6h}{h(2x + 2h - 1)(2x - 1)} = \frac{-6}{(2x + 2h - 1)(2x - 1)}
    • f′(x)=lim⁡h→0−6(2x+2h−1)(2x−1)=−6(2x−1)2f'(x) = \lim_{h \to 0} \frac{-6}{(2x + 2h - 1)(2x - 1)} = \frac{-6}{(2x - 1)^2}
    • Step 2: Find the slope of the tangent line at x=1x = 1.
    • f′(1)=−6(2(1)−1)2=−6(1)2=−6f'(1) = \frac{-6}{(2(1) - 1)^2} = \frac{-6}{(1)^2} = -6
    • Step 3: Find the y-coordinate at x=1x = 1.
    • f(1)=32(1)−1=31=3f(1) = \frac{3}{2(1) - 1} = \frac{3}{1} = 3
    • Step 4: Write the equation of the tangent line (point-slope form).
    • y−y<em>1=m(x−x</em>1)  ⟹  y−3=−6(x−1)y - y<em>1 = m(x - x</em>1) \implies y - 3 = -6(x - 1)
    • y=−6x+6+3  ⟹  y=−6x+9y = -6x + 6 + 3 \implies y = -6x + 9
  • ## Example: Finding the Derivative of a Square Root Function
    • Problem: Find f′(x)f'(x) if f(x)=1−3xf(x) = \sqrt{1 - 3x}.
    • Step 1: Form the difference quotient.
    • racf(x+h)−f(x)h=1−3(x+h)−1−3xh=1−3x−3h−1−3xhrac{f(x + h) - f(x)}{h} = \frac{\sqrt{1 - 3(x + h)} - \sqrt{1 - 3x}}{h} = \frac{\sqrt{1 - 3x - 3h} - \sqrt{1 - 3x}}{h}
    • Step 2: Multiply by the conjugate.
    • =1−3x−3h−1−3xh⋅1−3x−3h+1−3x1−3x−3h+1−3x= \frac{\sqrt{1 - 3x - 3h} - \sqrt{1 - 3x}}{h} \cdot \frac{\sqrt{1 - 3x - 3h} + \sqrt{1 - 3x}}{\sqrt{1 - 3x - 3h} + \sqrt{1 - 3x}}
    • =(1−3x−3h)−(1−3x)h(1−3x−3h+1−3x)=−3hh(1−3x−3h+1−3x)= \frac{(1 - 3x - 3h) - (1 - 3x)}{h(\sqrt{1 - 3x - 3h} + \sqrt{1 - 3x})} = \frac{-3h}{h(\sqrt{1 - 3x - 3h} + \sqrt{1 - 3x})}
    • =−31−3x−3h+1−3x= \frac{-3}{\sqrt{1 - 3x - 3h} + \sqrt{1 - 3x}}
    • Step 3: Take the limit as h→0h \to 0.
    • f′(x)=lim⁡h→0−31−3x−3h+1−3x=−321−3xf'(x) = \lim_{h \to 0} \frac{-3}{\sqrt{1 - 3x - 3h} + \sqrt{1 - 3x}} = \frac{-3}{2\sqrt{1 - 3x}}
  • ## Example: Finding the Velocity Function from a Position Function
    • Problem: If the position function of an object is given by f(t)=3t−t2f(t) = 3t - t^2, find the velocity function.
    • Solution: The velocity function, v(t)v(t), is the derivative of the position function, f′(t)f'(t).
    • Step 1: Form the difference quotient.
    • f(t+h)−f(t)=(3(t+h)−(t+h)2)−(3t−t2)f(t + h) - f(t) = (3(t + h) - (t + h)^2) - (3t - t^2)
    • =(3t+3h−(t2+2th+h2))−(3t−t2)= (3t + 3h - (t^2 + 2th + h^2)) - (3t - t^2)
    • =3t+3h−t2−2th−h2−3t+t2=3h−2th−h2= 3t + 3h - t^2 - 2th - h^2 - 3t + t^2 = 3h - 2th - h^2
    • racf(t+h)−f(t)h=3h−2th−h2h=3−2t−hrac{f(t + h) - f(t)}{h} = \frac{3h - 2th - h^2}{h} = 3 - 2t - h
    • Step 2: Take the limit as h→0h \to 0.
    • v(t)=f′(t)=lim⁡h→0(3−2t−h)=3−2tv(t) = f'(t) = \lim_{h \to 0} (3 - 2t - h) = 3 - 2t
  • ## Notations for the Derivative of a Function
    • If y=f(x)y = f(x), the following notations denote the derivative of ff:
    • f′(x)f'(x)
    • y′y'
    • dydx\frac{dy}{dx}
  • ## Differentiation
    • The process of taking the derivative of a function.
  • ## Differentiability at a Point
    • A function is differentiable at aa if f′(a)f'(a) exists.
  • ## Conditions Where a Function is NOT Differentiable
    • A function is not differentiable at the following types of locations:
    • Discontinuities: Points where the function has a break (e.g., jump, hole, vertical asymptote).
    • Corners: Points where the graph abruptly changes direction, resulting in different slopes from the left and right (e.g., absolute value functions).
    • Vertical Tangent Lines: Points where the tangent line is vertical, meaning its slope is undefined (e.g., x=0x = 0 for f(x)=x1/3f(x) = x^{1/3}).
  • ## Example: Identifying Non-Differentiable Points from a Graph
    • Based on the provided graph, the function is not differentiable at the following approximate locations:
    • x=−4x = -4 (Discontinuity - jump)
    • x=−2x = -2 (Corner)
    • x=2x = 2 (Corner)
    • x=6x = 6 (Discontinuity - hole)
    • x=8x = 8 (Vertical Tangent Line)
  • ## Example: Finding Non-Differentiable Points for an Absolute Value Function
    • Problem: Where is the function f(x)=∣x2−9∣f(x) = |x^2 - 9| not differentiable?
    • Solution: Absolute value functions typically create corners where their argument is zero.
    • Set the argument to zero: x2−9=0x^2 - 9 = 0
    • Factor: (x−3)(x+3)=0(x - 3)(x + 3) = 0
    • Therefore, x=3x = 3 and x=−3x = -3. At these points, the graph has sharp corners, making it non-differentiable.
  • ## Sketching the Graph of the Derivative from the Original Function
    • Procedure for sketching f′f' from the graph of ff:
    • Analyze the slope of f(x)f(x). The value of f′(x)f'(x) at any point xx is equal to the slope of the tangent line to f(x)f(x) at that point.
    • Positive Slope of ff: If f(x)f(x) is increasing, then f′(x)f'(x) will be positive (above the x-axis).
    • Negative Slope of ff: If f(x)f(x) is decreasing, then f′(x)f'(x) will be negative (below the x-axis).
    • Zero Slope of ff: If f(x)f(x) has a horizontal tangent (e.g., at local maxima or minima), then f′(x)f'(x) will be zero (crosses the x-axis).
    • Steepness of ff: The steeper the slope of f(x)f(x), the larger the absolute value of f′(x)f'(x). If f(x)f(x) is very steep and increasing, f′(x)f'(x) will be a large positive value. If very steep and decreasing, f′(x)f'(x) will be a large negative value.
    • Concavity of ff (or slope of f′f'): If f(x)f(x) is concave up, f′(x)f'(x) is increasing. If f(x)f(x) is concave down, f′(x)f'(x) is decreasing.
    • Points of Non-Differentiability in ff: If f(x)f(x) has a discontinuity, a corner, or a vertical tangent line, then f′(x)f'(x) will not exist at that corresponding x-value (often represented by a jump or a break in the graph of f′f').