Derivatives of Exponential and Logarithm Functions for MATH 122
Exam Relevance for MATH 122
Tested within chain rule and product/quotient rule problems. Know the basic forms.
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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