Derivatives of Exponential Functions

Exponential Functions

  • Definition: An exponential function has the form y=a(c)xy = a(c)^x, where (c) affects growth or decay.
    • Growth: (c > 1)
    • Decay: (0 < c < 1)
  • Alternate form used: y=k⋅axy = k \cdot a^x

Graph Characteristics

  • Example Function: f(x)=2xf(x) = 2^x
    • For (x < 0), function increases slowly; thus, f′f' values are small.
    • For (x > 0), function increases quickly; hence, f′f' values are larger.
  • Always concave up and increasing for all (x).
  • Expected Type of Derivative Function: Exponential

Tangent Line Calculation

  • Method: Calculate slopes of tangent lines at selected points.
  • Example Estimates (using graph):
    • At (x = -1), f′(−1)≈0.35f'(-1) \approx 0.35
    • At (x = 0), f′(0)≈0.7f'(0) \approx 0.7
    • At (x = 1), f′(1)≈1.4f'(1) \approx 1.4
    • At (x = 2), f′(2)≈2.75f'(2) \approx 2.75
    • At (x = 3), f′(3)≈5.5f'(3) \approx 5.5

Numerical Estimation of Derivative

  • Estimate f′(0)f'(0):
    • Numerical limit: limh→0f(x+h)−f(x)hlim_{h \to 0} \frac{f(x + h) - f(x)}{h}
    • Calculation at (x = 0):
    • Derivative defined as:
      f′(0)=lim<em>h→020+h−20h=lim</em>h→02h−1hf'(0) = lim<em>{h \to 0} \frac{2^{0+h} - 2^0}{h}= lim</em>{h \to 0} \frac{2^h - 1}{h}
    • Results at small (h):
      • At (h = -0.0003): 0.69310.6931
      • Approximate conclusion: f′(0)≈0.693f'(0) \approx 0.693

Patterns in Derivatives of Exponential Functions

  • Observations:
    • For f(x)=2xf(x) = 2^x, f′(x)≈0.693⋅2xf'(x) \approx 0.693 \cdot 2^x
    • Generalization: For any base, f′(x)=f′(0)⋅axf'(x) = f'(0) \cdot a^x.
  • Sample Function: g(x)=3xg(x) = 3^x
    • Based on calculations, g′(0)≈1.0986⋅3xg'(0) \approx 1.0986 \cdot 3^x

Derivative Relationships

  • Functions whose derivative is proportional to themselves:
    • f(x)=ax,f′(x)=kf(x).f(x) = a^x, f'(x) = k f(x).
    • Value of aa is crucial; investigatory values lead to base ee (approximately 2.718) as the only constant resulting in f′(x)=f(x)f'(x) = f(x).
    • Findings show that at zeros of their derivative function leads to:
      limh→0ah−1h=klim_{h \to 0}\frac{a^h - 1}{h} = k

Evaluating the Function y=exy = e^x

  • Features:
    • Zero x-intercepts
    • y-intercept: When (x = 0), y=1y = 1.
    • Domain: x∣x∈R{x | x \in \mathbb{R}}
    • Range: {y | y > 0, y \in \mathbb{R}}
    • Reflecting its inverse function: y=ln(x)y = ln(x).

Logarithm Review

  • Basic relationship: y=loga(x)⇔ay=xy = log_a(x)\Leftrightarrow a^y = x
  • Natural Logarithm: ln(x)ln(x), where base is ee.
  • ln and e are inverses.

Laws of Logarithms (base e)

  • Product Law: ln(MN)=ln(M)+ln(N)ln(MN) = ln(M) + ln(N)
  • Quotient Law: ln(MN)=ln(M)−ln(N)ln(\frac{M}{N}) = ln(M) - ln(N)
  • Power Law: ln(Mp)=p⋅ln(M)ln(M^p) = p \cdot ln(M)

Homework Problems

  • Review exercises and examples provided in the transcript for practice.