Premed · Premed · Calculus 1
Lecture 8: Derivatives of Exponential and Logarithmic Functions
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- State and apply the derivative of e^x and the general exponential a^x
- State and apply the derivative of ln x and the general logarithm log_a(x)
- Use logarithmic differentiation to handle complicated products, quotients, and variable exponents
- Differentiate functions of the form f(x)^{g(x)}
- Apply these derivatives in growth and decay models
Lecture Content
I. The Natural Exponential Function
The number e is defined as e = lim_{n -> infinity} (1 + 1/n)^n, which is approximately 2.71828. An equivalent characterization is that e is the unique number for which lim_{h -> 0} (e^h - 1)/h = 1. The derivative of e^x is d/dx [e^x] = e^x. The natural exponential is its own derivative, and this remarkable property is precisely what makes e special among all possible bases.
The proof follows directly from the definition: lim_{h -> 0} [e^{x+h} - e^x]/h = e^x lim_{h -> 0} (e^h - 1)/h = e^x 1 = e^x. When the chain rule is needed, d/dx [e^{g(x)}] = e^{g(x)} g'(x). For instance, d/dx [e^{3x^2}] = e^{3x^2} 6x = 6x * e^{3x^2}.
II. General Exponential Functions
For any base a > 0 with a not equal to 1, the exponential a^x can be rewritten as e^{x ln a}. Applying the chain rule to this form gives the derivative d/dx [a^x] = a^x ln a. When a = e, we recover d/dx [e^x] = e^x since ln e = 1. With a composite argument, the formula becomes d/dx [a^{g(x)}] = a^{g(x)} ln a g'(x). For example, d/dx [2^{sin x}] = 2^{sin x} ln 2 * cos x.
III. The Natural Logarithm
The natural logarithm ln x is the inverse of e^x and is defined for x > 0. Its derivative is d/dx [ln x] = 1/x. This can be proved by implicit differentiation: if y = ln x, then e^y = x. Differentiating both sides gives e^y * (dy/dx) = 1, so dy/dx = 1/e^y = 1/x.
With the chain rule, d/dx [ln(g(x))] = g'(x)/g(x), a formula that is extremely useful and appears frequently throughout calculus. For example, d/dx [ln(x^2 + 1)] = 2x/(x^2 + 1). When the argument may be negative, the formula d/dx [ln|x|] = 1/x holds for all x not equal to 0.
<image>Panel A: The graph of y = e^x and its tangent line at x = 0, showing that the slope equals the function value (slope = e^0 = 1 at x = 0, slope = e^1 = e at x = 1). Several tangent lines drawn at x = -1, 0, 1, 2, each with slope matching the y-coordinate. Panel B: The graph of y = ln x with tangent lines at x = 1 (slope 1), x = 2 (slope 1/2), and x = 4 (slope 1/4), illustrating that the slope equals 1/x. Title: "The exponential is its own derivative; the logarithm's derivative is the reciprocal."</image>
IV. General Logarithmic Functions
For any base a > 0 with a not equal to 1, the change-of-base formula gives log_a(x) = ln x / ln a. Its derivative is therefore d/dx [log_a(x)] = 1/(x ln a), which is just the natural logarithm derivative scaled by the constant 1/ln a. For example, d/dx [log_{10}(x)] = 1/(x ln 10).
V. Logarithmic Differentiation
Logarithmic differentiation is a technique for handling functions that are complicated products and quotients, or functions of the form f(x)^{g(x)} where both the base and exponent depend on x. The procedure has four steps: first, take the natural logarithm of both sides, so ln y = ln[expression]. Second, use logarithm properties to simplify, applying rules such as ln(AB) = ln A + ln B, ln(A/B) = ln A - ln B, and ln(A^n) = n ln A. Third, differentiate both sides implicitly, noting that the left side gives y'/y. Fourth, solve for y' and substitute back the original expression for y.
For example, to differentiate y = (x^3 * sqrt(x + 1)) / (x^2 + 2)^4, take logarithms to get ln y = 3 ln x + (1/2) ln(x + 1) - 4 ln(x^2 + 2). Differentiating implicitly gives y'/y = 3/x + 1/(2(x + 1)) - 8x/(x^2 + 2), and multiplying both sides by y yields the derivative.
<image>A step-by-step worked example of logarithmic differentiation for y = x^x. Step 1: Take ln of both sides — ln y = x ln x. Step 2: Differentiate implicitly — y'/y = ln x + x*(1/x) = ln x + 1. Step 3: Solve for y' — y' = y(ln x + 1) = x^x(ln x + 1). Each step is shown in a separate box with an arrow flowing downward. Title: "Logarithmic Differentiation: Finding the derivative of x^x."</image>
VI. Functions of the Form f(x)^{g(x)}
Functions with both a variable base and a variable exponent cannot be differentiated using the power rule (which requires a constant exponent) or the exponential rule (which requires a constant base). The approach is to rewrite the function as e^{g(x) ln f(x)} and differentiate using the chain rule, or equivalently, to use logarithmic differentiation.
For example, to differentiate x^{sin x}, rewrite it as e^{sin x ln x}. The derivative is e^{sin x ln x} d/dx [sin x ln x] = x^{sin x} [cos x ln x + sin x / x].
VII. Applications: Exponential Growth and Decay
If a quantity Q(t) satisfies the differential equation dQ/dt = kQ, then Q(t) = Q_0 * e^{kt}, where Q_0 is the initial value. When k > 0, this describes exponential growth, as seen in population models and compound interest. When k < 0, it describes exponential decay, as in radioactive decay and drug clearance from the body.
The half-life T_{1/2} is the time for Q to decrease to half its initial value. Setting Q_0 e^{k T_{1/2}} = Q_0 / 2 and solving gives T_{1/2} = ln 2 / |k|. The doubling time T_d = ln 2 / k describes how long exponential growth takes to double the initial quantity. These models are central in pharmacology, biology, and physics.
<image>Panel A: A graph of exponential growth Q(t) = Q_0 e^{kt} (k > 0) with the doubling time marked — showing Q_0 at t = 0 and 2Q_0 at t = T_d. Panel B: A graph of exponential decay Q(t) = Q_0 * e^{kt} (k < 0) with the half-life marked — showing Q_0 at t = 0 and Q_0/2 at t = T_{1/2}. Both curves approach the x-axis or grow unboundedly, with tangent lines drawn at t = 0 showing the initial rate of change. Title: "Exponential Growth and Decay."</image>
