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High school∫Calculus15 min read
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Exponential Function: Complete Lesson with Properties, Derivative and Interactive Graph

$e^x$ from A to Z: definition, algebraic properties, derivative, monotonicity, limits, equations and inequalities. With interactive graph.

Prerequisites :

  • Derivatives
  • Limits
  • Powers

The big picture

The exponential function, written exp⁑\exp or x↦exx \mapsto e^x, is the most important function in calculus. It has a unique property: it equals its own derivative.

-3-2-1123-1123456789101112(0, 1)(1, e)f(x) = e^x

Definition

exp⁑:Rβ†’R+βˆ—\exp : \mathbb{R} \to \mathbb{R}_+^*

The unique function on R\mathbb{R} such that:

  • exp⁑(0)=1\exp(0) = 1
  • For all real xx, exp⁑′(x)=exp⁑(x)\exp'(x) = \exp(x) (derivative equals the function itself)

Also written exp⁑(x)=ex\exp(x) = e^x, where eβ‰ˆ2.71828…e \approx 2.71828\ldots

The 5 essential algebraic properties

These look like power rules β€” they come up all the time:

PropertyFormula
Sum in exponentea+b=eaΓ—ebe^{a + b} = e^a \times e^b
Differenceeaβˆ’b=eaebe^{a - b} = \dfrac{e^a}{e^b}
Oppositeeβˆ’a=1eae^{-a} = \dfrac{1}{e^a}
Power(ea)n=ena(e^a)^n = e^{na}
Rootea/2=eae^{a/2} = \sqrt{e^a}

Derivative

Fundamental: (ex)β€²=ex(e^x)' = e^x

General (chain rule): (eu(x))β€²=uβ€²(x)β‹…eu(x)(e^{u(x)})' = u'(x) \cdot e^{u(x)}

Move the slider below: at every xx, the tangent slope equals the function’s value.

Point
β€”
Pente de la tangente (nombre dΓ©rivΓ©)
β€”

Examples:

  • (e3x)β€²=3e3x(e^{3x})' = 3 e^{3x}
  • (eβˆ’x)β€²=βˆ’eβˆ’x(e^{-x})' = -e^{-x}
  • (ex2)β€²=2xβ‹…ex2(e^{x^2})' = 2x \cdot e^{x^2}

Monotonicity

For all x∈Rx \in \mathbb{R}, ex>0e^x > 0 (exponential is always strictly positive).

So (ex)β€²=ex>0(e^x)' = e^x > 0 β‡’ exp⁑\exp is strictly increasing on R\mathbb{R}.

Consequence: exp⁑\exp is bijective from R\mathbb{R} to R+βˆ—\mathbb{R}_+^*. Its inverse is the natural log ln⁑\ln.

Limits to memorize

LimitValue
lim⁑xβ†’+∞ex\lim_{x \to +\infty} e^x+∞+\infty
lim⁑xβ†’βˆ’βˆžex\lim_{x \to -\infty} e^x00
lim⁑xβ†’+∞exx\lim_{x \to +\infty} \dfrac{e^x}{x}+∞+\infty (comparative growth)
lim⁑xβ†’+∞exxn\lim_{x \to +\infty} \dfrac{e^x}{x^n}+∞+\infty for all nn
lim⁑xβ†’βˆ’βˆžxβ‹…ex\lim_{x \to -\infty} x \cdot e^x00
lim⁑xβ†’0exβˆ’1x\lim_{x \to 0} \dfrac{e^x - 1}{x}11 (derivative at 0)

Comparative growth: exponential beats any power at +∞+\infty.

Equations and inequalities

Since exp⁑\exp is strictly increasing:

  • ea=ebβ€…β€ŠβŸΊβ€…β€Ša=be^a = e^b \iff a = b
  • ea<ebβ€…β€ŠβŸΊβ€…β€Ša<be^a < e^b \iff a < b

Example 1: equation

Solve e2xβˆ’3=ex+1e^{2x - 3} = e^{x + 1}.

By bijection: 2xβˆ’3=x+12x - 3 = x + 1, so x=4x = 4.

Example 2: with a constant

Solve ex=5e^x = 5.

Using ln⁑\ln: x=ln⁑(5)β‰ˆ1.61x = \ln(5) \approx 1.61.

Example 3: inequality

Solve e2xβˆ’1β‰₯e3x+2e^{2x - 1} \geq e^{3x + 2}.

By strict growth: 2xβˆ’1β‰₯3x+2β€…β€ŠβŸΊβ€…β€Šxβ‰€βˆ’32x - 1 \geq 3x + 2 \iff x \leq -3.

Full study: f(x)=xeβˆ’xf(x) = x e^{-x}

A classic exam question.

Domain: R\mathbb{R}.

Derivative: fβ€²(x)=eβˆ’x(1βˆ’x)f'(x) = e^{-x}(1 - x).

Sign of fβ€²f': since eβˆ’x>0e^{-x} > 0, sign of (1βˆ’x)(1 - x):

  • fβ€²>0f' > 0 for x<1x < 1
  • fβ€²<0f' < 0 for x>1x > 1

Extremum: max at x=1x = 1, f(1)=1/eβ‰ˆ0.37f(1) = 1/e \approx 0.37.

Limits:

  • As xβ†’βˆ’βˆžx \to -\infty: eβˆ’xβ†’+∞e^{-x} \to +\infty and xβ†’βˆ’βˆžx \to -\infty, so f(x)β†’βˆ’βˆžf(x) \to -\infty.
  • As xβ†’+∞x \to +\infty: x/exβ†’0x/e^x \to 0.
-2-112345-11maxf(x) = xΒ·e^(-x)

Real-world applications

  • Population growth: N(t)=N0β‹…ektN(t) = N_0 \cdot e^{kt}
  • Radioactive decay: N(t)=N0β‹…eβˆ’Ξ»tN(t) = N_0 \cdot e^{-\lambda t}
  • Capacitor charging: u(t)=U(1βˆ’eβˆ’t/Ο„)u(t) = U(1 - e^{-t/\tau})
  • Continuous compound interest: C(t)=C0β‹…ertC(t) = C_0 \cdot e^{rt}

Common mistakes

  1. Confusing ea+be^{a+b} with ea+ebe^a + e^b: it’s a product (eaΓ—ebe^a \times e^b), not a sum.
  2. Forgetting uβ€²u' in chain rule: (e2x)β€²=2e2x(e^{2x})' = 2 e^{2x}, NOT e2xe^{2x}.
  3. Thinking eβˆ’xe^{-x} can be negative: NO, exponential is always > 0.
  4. ln⁑\ln and ee are inverses: ln⁑(ex)=x\ln(e^x) = x and eln⁑(x)=xe^{\ln(x)} = x (for x>0x > 0).

Solved exercises

Exercise 1

Simplify e2xβ‹…eβˆ’3xe^{2x} \cdot e^{-3x}.

Show solution

Sum exponents: eβˆ’xe^{-x}.

Exercise 2

Solve ex2=e2x+3e^{x^2} = e^{2x + 3}.

Show solution

By bijection: x2=2x+3x^2 = 2x + 3, so x2βˆ’2xβˆ’3=0x^2 - 2x - 3 = 0. Ξ”=16\Delta = 16, roots βˆ’1-1 and 33.

Exercise 3

Study monotonicity of g(x)=ex2βˆ’2xg(x) = e^{x^2 - 2x}.

Show solution

gβ€²(x)=(2xβˆ’2)ex2βˆ’2xg'(x) = (2x - 2) e^{x^2 - 2x}. Since e...>0e^{...} > 0:

  • gβ€²<0g' < 0 for x<1x < 1
  • gβ€²>0g' > 0 for x>1x > 1

gg decreases on (βˆ’βˆž,1](-\infty, 1], increases on [1,+∞)[1, +\infty). Min at x=1x = 1: g(1)=1/eg(1) = 1/e.

Takeaways

  • exp⁑(0)=1\exp(0) = 1, exp⁑′=exp⁑\exp' = \exp.
  • Always strictly positive, strictly increasing, bijective.
  • ea+b=eaebe^{a+b} = e^a e^b, eβˆ’a=1/eae^{-a} = 1/e^a.
  • Chain rule: (eu)β€²=uβ€²eu(e^u)' = u' e^u.
  • Beats any power at +∞+\infty.
  • Inverse: ln⁑\ln.

Going further

#exponential#e^x#grade 12#AP calc#lesson