The derivative of $e^x$ is itself! This makes $e$ the natural choice for calculus.

💡 Amazing Fact: $e^x$ is the ONLY function (up to a constant multiple) that is its own derivative!

Why e is Special

The number $e \approx 2.71828...$ is called Euler's number, and it's defined specifically so that:

$\lim_{h \to 0} \frac{e^h - 1}{h} = 1$

This property makes the derivative formula so clean!

Exponential Functions with Chain Rule

When the exponent is not just $x$, use the Chain Rule:

General Formula

$\frac{d}{dx}[e^u] = e^u \cdot u'$

where $u$ is any function of $x$.

Examples

1. $\frac{d}{dx}[e^{2x}] = e^{2x} \cdot 2 = 2e^{2x}$

2. $\frac{d}{dx}[e^{x^2}] = e^{x^2} \cdot 2x = 2xe^{x^2}$

3. $\frac{d}{dx}[e^{-3x}] = e^{-3x} \cdot (-3) = -3e^{-3x}$

4. $\frac{d}{dx}[e^{\sin x}] = e^{\sin x} \cdot \cos x = e^{\sin x}\cos x$

💡 Pattern: The exponential part stays the same, just multiply by the derivative of the exponent!

General Exponential Functions: a^x

For exponential functions with other bases:

$\frac{d}{dx}[a^x] = a^x \ln a$

Examples

1. $\frac{d}{dx}[2^x] = 2^x \ln 2$

2. $\frac{d}{dx}[10^x] = 10^x \ln 10$

3. $\frac{d}{dx}[3^{2x}] = 3^{2x} \ln 3 \cdot 2 = 2\ln(3) \cdot 3^{2x}$

Why the ln a?

When $a = e$, we get $\ln e = 1$, so the formula becomes $\frac{d}{dx}[e^x] = e^x \cdot 1 = e^x$ ✓

Combining Rules

Exponential derivatives often require multiple rules:

Product Rule + Exponential

Example: $f(x) = xe^x$

Using $u = x$ and $v = e^x$:

$f'(x) = (1)(e^x) + (x)(e^x) = e^x + xe^x = e^x(1 + x)$

Quotient Rule + Exponential

Example: $g(x) = \frac{e^x}{x}$

Using the quotient rule:

$g'(x) = \frac{(e^x)(x) - (e^x)(1)}{x^2} = \frac{xe^x - e^x}{x^2} = \frac{e^x(x-1)}{x^2}$

Chain Rule + Exponential + Trig

Example: $h(x) = e^{\sin(2x)}$

$h'(x) = e^{\sin(2x)} \cdot \cos(2x) \cdot 2 = 2\cos(2x)e^{\sin(2x)}$

Applications

Population Growth

If $P(t) = 1000e^{0.05t}$ represents a population:

$P'(t) = 1000e^{0.05t} \cdot 0.05 = 50e^{0.05t}$

This gives the rate of growth at time $t$.

Radioactive Decay

If $N(t) = N_0 e^{-\lambda t}$ represents radioactive atoms:

$N'(t) = N_0 e^{-\lambda t} \cdot (-\lambda) = -\lambda N_0 e^{-\lambda t}$

The negative sign shows the quantity is decreasing.

Compound Interest

If $A(t) = 5000e^{0.06t}$ represents account balance:

$A'(t) = 5000e^{0.06t} \cdot 0.06 = 300e^{0.06t}$

This is the instantaneous rate at which money is being earned.

⚠️ Common Mistakes

Mistake 1: Forgetting Chain Rule

❌ $\frac{d}{dx}[e^{2x}] = e^{2x}$ ✅ $\frac{d}{dx}[e^{2x}] = 2e^{2x}$

Mistake 2: Confusing with Power Rule

❌ $\frac{d}{dx}[e^x] = xe^{x-1}$ (This is WRONG! Not a power of $x$!) ✅ $\frac{d}{dx}[e^x] = e^x$

Mistake 3: Wrong Base Formula

❌ $\frac{d}{dx}[2^x] = 2^x$ ✅ $\frac{d}{dx}[2^x] = 2^x \ln 2$

Mistake 4: Algebraic Errors

$e^{x+1} = e^x \cdot e^1 = e \cdot e^x$, NOT $e^x + e$

Special Cases and Tricks

Constants in Exponents

$\frac{d}{dx}[e^{3x+5}] = e^{3x+5} \cdot 3 = 3e^{3x+5}$

The constant 5 disappears when differentiating!

Negative Exponents

$\frac{d}{dx}[e^{-x}] = e^{-x} \cdot (-1) = -e^{-x}$

Products with e^x Factor Out!

$xe^x + e^x = e^x(x + 1)$ — Always factor out $e^x$ when possible!

📝 Key Formulas to Memorize

1. $\frac{d}{dx}[e^x] = e^x$

2. $\frac{d}{dx}[e^u] = e^u \cdot u'$ (Chain Rule)

3. $\frac{d}{dx}[a^x] = a^x \ln a$ (General base)

4. $\frac{d}{dx}[a^u] = a^u \ln a \cdot u'$ (General base + Chain Rule)

Practice Strategy

1. Identify if it's base $e$ or another base
2. Look for what's in the exponent
3. If exponent is not just $x$, prepare to use Chain Rule
4. Apply the formula and multiply by derivative of exponent
5. Factor out $e^x$ or $a^x$ when simplifying

Step 2: Find $u'$

$u' = 2x$

Step 3: Find $v'$ (using Chain Rule)

$v' = e^{3x} \cdot 3 = 3e^{3x}$

Step 4: Apply Product Rule

$\frac{dy}{dx} = u'v + uv'$

$= (2x)(e^{3x}) + (x^2)(3e^{3x})$

$= 2xe^{3x} + 3x^2e^{3x}$

Step 5: Factor out $e^{3x}$

$= e^{3x}(2x + 3x^2)$

$= e^{3x}(3x^2 + 2x)$

$= xe^{3x}(3x + 2)$

Answer: $\displaystyle\frac{dy}{dx} = xe^{3x}(3x + 2)$ or $e^{3x}(3x^2 + 2x)$

Step 2: Find $u'$ (Chain Rule)

$u' = e^{2x} \cdot 2 = 2e^{2x}$

Step 3: Find $v'$

$v' = 3x^2$

Step 4: Apply Quotient Rule

$g'(x) = \frac{u'v - uv'}{v^2}$

$= \frac{(2e^{2x})(x^3) - (e^{2x})(3x^2)}{(x^3)^2}$

$= \frac{2x^3e^{2x} - 3x^2e^{2x}}{x^6}$

Step 5: Factor out common terms

Factor out $e^{2x}$ and $x^2$:

$= \frac{x^2e^{2x}(2x - 3)}{x^6}$

$= \frac{e^{2x}(2x - 3)}{x^4}$

Answer: $\displaystyle g'(x) = \frac{e^{2x}(2x - 3)}{x^4}$