Derivatives of Exponential and Logarithm Functions for MATH 122
Exam Relevance for MATH 122
Exponential and log derivatives are heavily tested in MATH 122 due to business applications like compound interest.
Derivatives of Exponential Functions
The Natural Exponential: $e^x$
The function $e^x$ is special because it is its own derivative:
$$\frac{d}{dx}[e^x] = e^x$$
This is the only function (other than 0) with this property!
Problem: Find $\frac{d}{dx}[e^x]$
Step 1: Apply the rule $\frac{d}{dx}[e^x] = e^x$:
$$\boxed{e^x}$$
Problem: Find $\frac{d}{dx}[e^{3x}]$
Step 1: Identify this as $e^{f(x)}$ where $f(x) = 3x$
Step 2: Apply the chain rule: $\frac{d}{dx}[e^{f(x)}] = e^{f(x)} \cdot f'(x)$
$$\frac{d}{dx}[e^{3x}] = e^{3x} \cdot 3 = \boxed{3e^{3x}}$$
Problem: Find $\frac{d}{dx}[e^{x^2}]$
Step 1: Here $f(x) = x^2$, so $f'(x) = 2x$
Step 2: Apply chain rule: $$\frac{d}{dx}[e^{x^2}] = e^{x^2} \cdot 2x = \boxed{2xe^{x^2}}$$
General Exponential: $a^x$
For any positive base $a \neq 1$:
$$\frac{d}{dx}[a^x] = a^x \ln(a)$$
Why the $\ln(a)$? We can rewrite $a^x = e^{x\ln(a)}$, then use chain rule.
Problem: Find $\frac{d}{dx}[2^x]$
Step 1: Apply the rule $\frac{d}{dx}[a^x] = a^x \ln(a)$ with $a = 2$: $$\frac{d}{dx}[2^x] = 2^x \cdot \ln(2)$$
$$\boxed{2^x \ln 2}$$
Derivatives of Logarithmic Functions
The Natural Logarithm: $\ln x$
$$\frac{d}{dx}[\ln x] = \frac{1}{x} \quad \text{(for } x > 0 \text{)}$$
Problem: Find $\frac{d}{dx}[\ln x]$
Step 1: Apply the rule $\frac{d}{dx}[\ln x] = \frac{1}{x}$:
$$\boxed{\frac{1}{x}}$$
Problem: Find $\frac{d}{dx}[\ln(3x+1)]$
Step 1: This is $\ln(f(x))$ where $f(x) = 3x+1$
Step 2: Apply chain rule: $\frac{d}{dx}[\ln f(x)] = \frac{f'(x)}{f(x)}$
$$\frac{d}{dx}[\ln(3x+1)] = \frac{3}{3x+1} = \boxed{\frac{3}{3x+1}}$$
Problem: Find $\frac{d}{dx}[\ln(x^2)]$
Solution Method 1: Use chain rule with $f(x) = x^2$: $$\frac{d}{dx}[\ln(x^2)] = \frac{2x}{x^2} = \frac{2}{x}$$
Method 2: Simplify first using log properties: $\ln(x^2) = 2\ln x$ $$\frac{d}{dx}[2\ln x] = 2 \cdot \frac{1}{x} = \frac{2}{x}$$
$$\boxed{\frac{2}{x}}$$
General Logarithm: $\log_a x$
$$\frac{d}{dx}[\log_a x] = \frac{1}{x \ln(a)}$$
Problem: Find $\frac{d}{dx}[\log_{10} x]$
Step 1: Apply the rule $\frac{d}{dx}[\log_a x] = \frac{1}{x \ln(a)}$ with $a = 10$: $$\frac{d}{dx}[\log_{10} x] = \frac{1}{x \ln(10)}$$
$$\boxed{\frac{1}{x \ln 10}}$$
Summary Table
| Function | Derivative |
|---|---|
| $e^x$ | $e^x$ |
| $e^{f(x)}$ | $e^{f(x)} \cdot f'(x)$ |
| $a^x$ | $a^x \ln(a)$ |
| $\ln x$ | $\frac{1}{x}$ |
| $\ln f(x)$ | $\frac{f'(x)}{f(x)}$ |
| $\log_a x$ | $\frac{1}{x \ln(a)}$ |
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting the chain rule with exponentials
Wrong: $\frac{d}{dx}[e^{5x}] = e^{5x}$
Why it's wrong: The exponent $5x$ is not just $x$, so you need the chain rule.
Correct: $\frac{d}{dx}[e^{5x}] = e^{5x} \cdot 5 = 5e^{5x}$
❌ Mistake: Confusing $\frac{d}{dx}[e^x]$ with $\frac{d}{dx}[x^e]$
- $\frac{d}{dx}[e^x] = e^x$ (exponential rule — base is constant)
- $\frac{d}{dx}[x^e] = ex^{e-1}$ (power rule — exponent is constant)
The rules depend on what's constant and what's variable!
Derivative of e^x
The exponential function e^x is its own derivative
Variables:
- $e$:
- Euler's number (≈ 2.718)
- $e^x$:
- the natural exponential function
Derivative of e^(f(x))
Chain rule applied to exponential functions
Variables:
- $f(x)$:
- the exponent function
- $f'(x)$:
- derivative of the exponent
Derivative of a^x
Derivative of exponential with any base a
Variables:
- $a$:
- the base (a > 0, a ≠ 1)
- $ln(a)$:
- natural logarithm of the base
Derivative of ln(x)
Derivative of the natural logarithm
Variables:
- $x$:
- must be positive (x > 0)
Derivative of ln(f(x))
Chain rule applied to natural logarithm
Variables:
- $f(x)$:
- the inner function (must be positive)
- $f'(x)$:
- derivative of the inner function
Derivative of log_a(x)
Derivative of logarithm with any base
Variables:
- $a$:
- the base of the logarithm
- $ln(a)$:
- natural log of the base
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