Derivatives of Exponential Functions

Finding derivatives involving e^x and other exponential functions

🚀 Derivatives of Exponential Functions

The Most Important Derivative: e^x

The exponential function with base ee has a remarkable property:

ddx[ex]=ex\frac{d}{dx}[e^x] = e^x

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

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


Why e is Special

The number e2.71828...e \approx 2.71828... is called Euler's number, and it's defined specifically so that:

limh0eh1h=1\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 xx, use the Chain Rule:

General Formula

ddx[eu]=euu\frac{d}{dx}[e^u] = e^u \cdot u'

where uu is any function of xx.

Examples

  1. ddx[e2x]=e2x2=2e2x\frac{d}{dx}[e^{2x}] = e^{2x} \cdot 2 = 2e^{2x}

  2. ddx[ex2]=ex22x=2xex2\frac{d}{dx}[e^{x^2}] = e^{x^2} \cdot 2x = 2xe^{x^2}

  3. ddx[e3x]=e3x(3)=3e3x\frac{d}{dx}[e^{-3x}] = e^{-3x} \cdot (-3) = -3e^{-3x}

  4. ddx[esinx]=esinxcosx=esinxcosx\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:

ddx[ax]=axlna\frac{d}{dx}[a^x] = a^x \ln a

Examples

  1. ddx[2x]=2xln2\frac{d}{dx}[2^x] = 2^x \ln 2

  2. ddx[10x]=10xln10\frac{d}{dx}[10^x] = 10^x \ln 10

  3. ddx[32x]=32xln32=2ln(3)32x\frac{d}{dx}[3^{2x}] = 3^{2x} \ln 3 \cdot 2 = 2\ln(3) \cdot 3^{2x}

Why the ln a?

When a=ea = e, we get lne=1\ln e = 1, so the formula becomes ddx[ex]=ex1=ex\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)=xexf(x) = xe^x

Using u=xu = x and v=exv = e^x:

f(x)=(1)(ex)+(x)(ex)=ex+xex=ex(1+x)f'(x) = (1)(e^x) + (x)(e^x) = e^x + xe^x = e^x(1 + x)

Quotient Rule + Exponential

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

Using the quotient rule:

g(x)=(ex)(x)(ex)(1)x2=xexexx2=ex(x1)x2g'(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)=esin(2x)h(x) = e^{\sin(2x)}

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


Applications

Population Growth

If P(t)=1000e0.05tP(t) = 1000e^{0.05t} represents a population:

P(t)=1000e0.05t0.05=50e0.05tP'(t) = 1000e^{0.05t} \cdot 0.05 = 50e^{0.05t}

This gives the rate of growth at time tt.

Radioactive Decay

If N(t)=N0eλtN(t) = N_0 e^{-\lambda t} represents radioactive atoms:

N(t)=N0eλt(λ)=λN0eλtN'(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)=5000e0.06tA(t) = 5000e^{0.06t} represents account balance:

A(t)=5000e0.06t0.06=300e0.06tA'(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

ddx[e2x]=e2x\frac{d}{dx}[e^{2x}] = e^{2x}ddx[e2x]=2e2x\frac{d}{dx}[e^{2x}] = 2e^{2x}

Mistake 2: Confusing with Power Rule

ddx[ex]=xex1\frac{d}{dx}[e^x] = xe^{x-1} (This is WRONG! Not a power of xx!) ✅ ddx[ex]=ex\frac{d}{dx}[e^x] = e^x

Mistake 3: Wrong Base Formula

ddx[2x]=2x\frac{d}{dx}[2^x] = 2^xddx[2x]=2xln2\frac{d}{dx}[2^x] = 2^x \ln 2

Mistake 4: Algebraic Errors

ex+1=exe1=eexe^{x+1} = e^x \cdot e^1 = e \cdot e^x, NOT ex+ee^x + e


Special Cases and Tricks

Constants in Exponents

ddx[e3x+5]=e3x+53=3e3x+5\frac{d}{dx}[e^{3x+5}] = e^{3x+5} \cdot 3 = 3e^{3x+5}

The constant 5 disappears when differentiating!

Negative Exponents

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

Products with e^x Factor Out!

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


📝 Key Formulas to Memorize

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

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

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

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


Practice Strategy

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

📚 Practice Problems

1Problem 1easy

Question:

Find the derivative of f(x)=3e4x+2exf(x) = 3e^{4x} + 2e^{-x}.

💡 Show Solution

Step 1: Apply the Sum Rule

Take the derivative of each term separately:

f(x)=ddx[3e4x]+ddx[2ex]f'(x) = \frac{d}{dx}[3e^{4x}] + \frac{d}{dx}[2e^{-x}]


Step 2: First term - use Chain Rule

ddx[3e4x]=3e4x4=12e4x\frac{d}{dx}[3e^{4x}] = 3 \cdot e^{4x} \cdot 4 = 12e^{4x}


Step 3: Second term - use Chain Rule

ddx[2ex]=2ex(1)=2ex\frac{d}{dx}[2e^{-x}] = 2 \cdot e^{-x} \cdot (-1) = -2e^{-x}


Step 4: Combine

f(x)=12e4x2exf'(x) = 12e^{4x} - 2e^{-x}

Answer: f(x)=12e4x2exf'(x) = 12e^{4x} - 2e^{-x}

2Problem 2medium

Question:

Find dydx\frac{dy}{dx} if y=x2e3xy = x^2e^{3x}.

💡 Show Solution

This requires the Product Rule.


Step 1: Identify the product

u=x2u = x^2 and v=e3xv = e^{3x}


Step 2: Find uu'

u=2xu' = 2x


Step 3: Find vv' (using Chain Rule)

v=e3x3=3e3xv' = e^{3x} \cdot 3 = 3e^{3x}


Step 4: Apply Product Rule

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

=(2x)(e3x)+(x2)(3e3x)= (2x)(e^{3x}) + (x^2)(3e^{3x})

=2xe3x+3x2e3x= 2xe^{3x} + 3x^2e^{3x}


Step 5: Factor out e3xe^{3x}

=e3x(2x+3x2)= e^{3x}(2x + 3x^2)

=e3x(3x2+2x)= e^{3x}(3x^2 + 2x)

=xe3x(3x+2)= xe^{3x}(3x + 2)

Answer: dydx=xe3x(3x+2)\displaystyle\frac{dy}{dx} = xe^{3x}(3x + 2) or e3x(3x2+2x)e^{3x}(3x^2 + 2x)

3Problem 3hard

Question:

Find the derivative of g(x)=e2xx3g(x) = \frac{e^{2x}}{x^3}.

💡 Show Solution

This requires the Quotient Rule.


Step 1: Identify numerator and denominator

u=e2xu = e^{2x} (top)

v=x3v = x^3 (bottom)


Step 2: Find uu' (Chain Rule)

u=e2x2=2e2xu' = e^{2x} \cdot 2 = 2e^{2x}


Step 3: Find vv'

v=3x2v' = 3x^2


Step 4: Apply Quotient Rule

g(x)=uvuvv2g'(x) = \frac{u'v - uv'}{v^2}

=(2e2x)(x3)(e2x)(3x2)(x3)2= \frac{(2e^{2x})(x^3) - (e^{2x})(3x^2)}{(x^3)^2}

=2x3e2x3x2e2xx6= \frac{2x^3e^{2x} - 3x^2e^{2x}}{x^6}


Step 5: Factor out common terms

Factor out e2xe^{2x} and x2x^2:

=x2e2x(2x3)x6= \frac{x^2e^{2x}(2x - 3)}{x^6}

=e2x(2x3)x4= \frac{e^{2x}(2x - 3)}{x^4}

Answer: g(x)=e2x(2x3)x4\displaystyle g'(x) = \frac{e^{2x}(2x - 3)}{x^4}