Derivatives

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Derivatives #

CalculusDifficulty: ★★☆☆☆Depth: 2Unlocks: 91

Instantaneous rate of change. Slope of tangent line.

Interactive Visualization #

⏮◀◀▶▶STEP0.25x1xZOOM

t=0s

Core Concepts #

Key Symbols & Notation #

f'(a) (derivative of f at a)

Essential Relationships #

Prerequisites (2) #

Limits5 atomsSlope and Rate of Change5 atoms

Unlocks (8) #

Derivative Ruleslvl 2Maximum Likelihood Estimationlvl 3Convex Functionslvl 3Integralslvl 2Utility Theorylvl 3Demand Functionslvl 3Cost Functionslvl 3Price Elasticitylvl 3

Referenced by (6) #

Where this concept shows up in the operating-finance and personal-finance graphs.

From Business (4) #

[Time to ValueBusiness

TTV is governed by the rate of value accrual over time; the derivative of cumulative value with respect to time formalizes whether value delivery is accelerating or stalling, which determines whether TTV is short or long](/business/time-to-value/)[DepreciationBusiness

The 'net rate' that determines investment type is literally a derivative - the instantaneous rate of change of asset value over time. Depreciation is a negative derivative, appreciation is positive, and the investment classification depends on their sum's sign.](/business/depreciation/)[overheadBusiness

Detecting overhead means measuring the marginal gain in downstream performance per added τ, which is literally evaluating a discrete derivative and checking when it approaches zero](/business/overhead/)[Book ValueBusiness

The first derivative of book value over time tells you whether it is compounding (positive, increasing) or depreciating (negative). The second derivative tells you if that trend is accelerating or decelerating - the mathematical formalism for 'is this asset gaining or losing value, and how fast?'](/business/book-value/)

From Money (2) #

[Return on EquityMoney

Marginal return on equity measures the sensitivity of returns to capital redeployment](/money/return-on-equity/)[Options BasicsMoney

Option pricing uses calculus derivatives for sensitivity analysis (the Greeks)](/money/options-basics/)

Advanced Learning Details

Graph Position #

28

Depth Cost

91

Fan-Out (ROI)

33

Bottleneck Score

2

Chain Length

Cognitive Load #

6

Atomic Elements

22

Total Elements

L0

Percentile Level

L4

Atomic Level

All Concepts (8) #

Teaching Strategy #

Deep-dive lesson - accessible entry point but dense material. Use worked examples and spaced repetition.

If you zoom in far enough on a smooth curve, it starts to look like a straight line. A derivative is the slope of that “best-fit” line at a single point—capturing an instantaneous rate of change.

TL;DR:

The derivative of fff at aaa, written f′(a)f'(a)f′(a), is the limit of slopes of secant lines through (a,f(a))(a,f(a))(a,f(a)) and nearby points. Formally:

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}f′(a)=h→0lim​hf(a+h)−f(a)​

If the limit exists, the function is differentiable at aaa and the derivative equals the slope of the tangent line there (and the instantaneous rate of change).

What Is a Derivative? #

The motivation (why we need a new idea) #

You already know average rate of change: between two points x=ax=ax=a and x=bx=bx=b, the slope is

f(b)−f(a)b−a\frac{f(b)-f(a)}{b-a}b−af(b)−f(a)​

That’s a two-point measurement. But many real questions are one-point questions:

To answer those, we take the average rate of change over a smaller and smaller interval and ask whether it approaches a limiting value.

Definition (difference quotient → limit) #

Fix a point aaa. Consider a nearby point a+ha+ha+h (so hhh is a small horizontal step). The slope of the secant line through

is

msec(h)=f(a+h)−f(a)(a+h)−a=f(a+h)−f(a)hm_{\text{sec}}(h)=\frac{f(a+h)-f(a)}{(a+h)-a}=\frac{f(a+h)-f(a)}{h}msec​(h)=(a+h)−af(a+h)−f(a)​=hf(a+h)−f(a)​

If this slope approaches a single number as h→0h \to 0h→0, we define the derivative:

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}f′(a)=h→0lim​hf(a+h)−f(a)​

This is the defining limit (also called the definition via the difference quotient).

Two interpretations you should hold at the same time #

  1. Geometric: f′(a)f'(a)f′(a) is the slope of the tangent line to the graph at x=ax=ax=a.

  2. Dynamic (rate): f′(a)f'(a)f′(a) is the instantaneous rate of change of fff with respect to xxx at x=ax=ax=a.

Both interpretations come from the same limit.

Units (a quick reality check) #

If xxx is measured in seconds and f(x)f(x)f(x) in meters, then

So f′(a)f'(a)f′(a) has the same units as a speed. This “units check” is one of the simplest ways to catch mistakes.

Supporting static diagram (to reduce cognitive load) #

Static diagram to include in the lesson UI: a coordinate plane with the curve y=f(x)y=f(x)y=f(x), points P=(a,f(a))P=(a,f(a))P=(a,f(a)) and Q=(a+h,f(a+h))Q=(a+h,f(a+h))Q=(a+h,f(a+h)), the secant line through PPP and QQQ, and labels for:

This diagram should remain visible while the definition is introduced so the symbols feel anchored to geometry.

What the interactive canvas should show (visual layer) #

Your interactive canvas should make the limit feel physical.

Canvas scene A: Secant-to-tangent animation

Canvas scene B: One-sided convergence

Canvas scene C: Non-differentiable corner/cusp toggle

These visuals directly support the three atomic concepts: instantaneous rate, defining limit, and tangent slope.

Core Mechanic 1: From Secant Slopes to the Tangent Slope (The Limit Process) #

Start with what you know: secant slope #

Pick two x-values: aaa and a+ha+ha+h.

The secant slope is

msec(h)=f(a+h)−f(a)hm_{\text{sec}}(h)=\frac{f(a+h)-f(a)}{h}msec​(h)=hf(a+h)−f(a)​

This is an average rate of change on the interval [a,a+h][a, a+h][a,a+h].

Shrink the interval (the key move) #

To get a one-point notion, we shrink the interval by taking hhh smaller and smaller.

The derivative is the number the secant slopes are heading toward—if they head toward a single number at all.

Why the limit is necessary (not optional) #

You might wonder: why not just set h=0h=0h=0 in

f(a+h)−f(a)h?\frac{f(a+h)-f(a)}{h}?hf(a+h)−f(a)​?

Because you would get

f(a)−f(a)0=00\frac{f(a)-f(a)}{0}=\frac{0}{0}0f(a)−f(a)​=00​

and 00\frac{0}{0}00​ is indeterminate (it doesn’t have a single value). The limit process asks: as we approach the problematic point, does the expression stabilize?

One-sided limits and differentiability #

A subtle but crucial point: h→0h \to 0h→0 means approaching 0 from both sides.

f+′(a)=lim⁡h→0+f(a+h)−f(a)hf'_+(a)=\lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}f+′​(a)=h→0+lim​hf(a+h)−f(a)​

f−′(a)=lim⁡h→0−f(a+h)−f(a)hf'_-(a)=\lim_{h\to 0^-}\frac{f(a+h)-f(a)}{h}f−′​(a)=h→0−lim​hf(a+h)−f(a)​

A function is differentiable at aaa only if both exist and are equal:

f′(a) exists   ⟺  f−′(a)=f+′(a)f'(a) \text{ exists } \iff f'_-(a)=f'_+(a)f′(a) exists ⟺f−′​(a)=f+′​(a)

This is exactly what your “one-sided slope convergence” canvas should make obvious: you can see when the two sides don’t agree.

Tangent line equation (once you have the slope) #

If you know f′(a)f'(a)f′(a), you know the slope of the tangent line at x=ax=ax=a. The tangent line passes through (a,f(a))(a, f(a))(a,f(a)) with slope f′(a)f'(a)f′(a):

y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x-a)y−f(a)=f′(a)(x−a)

This is often the fastest way to turn “derivative” into a concrete geometric object.

Micro-change viewpoint (a preview of linear approximation) #

When hhh is small, the change in fff is approximately

f(a+h)−f(a)≈f′(a) hf(a+h)-f(a) \approx f'(a),hf(a+h)−f(a)≈f′(a)h

This is the intuition behind many applications: derivatives translate a tiny input change into an approximate output change.

What the interactive canvas should emphasize here #

To reinforce the limit idea (instead of treating it as symbolism):

h(f(a+h) − f(a))/h
1...
0.5...
0.1...
0.01...

Core Mechanic 2: When Derivatives Do (and Don’t) Exist #

Differentiable vs. continuous (don’t merge these) #

A common mental trap is to assume:

Not always.

Facts to keep straight:

So differentiability is a stronger condition.

Three classic ways differentiability can fail #

1) Corner (left and right slopes disagree) #

Example: f(x)=∣x∣f(x)=|x|f(x)=∣x∣ at x=0x=0x=0.

Compute using the definition:

For h>0h>0h>0:

∣0+h∣−∣0∣h=∣h∣h=hh=1\frac{|0+h|-|0|}{h}=\frac{|h|}{h}=\frac{h}{h}=1h∣0+h∣−∣0∣​=h∣h∣​=hh​=1

For h<0h<0h<0:

∣h∣h=−hh=−1\frac{|h|}{h}=\frac{-h}{h}=-1h∣h∣​=h−h​=−1

So

f+′(0)=1,f−′(0)=−1f'_+(0)=1,\quad f'_-(0)=-1f+′​(0)=1,f−′​(0)=−1

They differ, so f′(0)f'(0)f′(0) does not exist.

Interactive tie-in: in your corner toggle, the secant line will approach two different tangent candidates depending on direction.

2) Cusp or vertical tangent (slope blows up) #

Example: f(x)=x3f(x)=\sqrt[3]{x}f(x)=3x​ at x=0x=0x=0.

Consider the difference quotient:

h3−0h=h1/3h=h−2/3\frac{\sqrt[3]{h}-0}{h}=\frac{h^{1/3}}{h}=h^{-2/3}h3h​−0​=hh1/3​=h−2/3

As h→0h\to 0h→0, h−2/3→∞h^{-2/3}\to \inftyh−2/3→∞. The slope becomes unbounded (vertical tangent). In many calculus courses, we say the derivative does not exist as a finite number.

Interactive tie-in: your slope readout should grow very large in magnitude as hhh shrinks.

3) Discontinuity (jumps or holes) #

If fff isn’t continuous at aaa, it cannot be differentiable there.

This is less subtle visually—your curve breaks—so it’s a good “sanity check” case.

Differentiability is local smoothness #

A useful phrase: differentiable at $a$ means the graph is locally well-approximated by a line near $a$.

That’s why the “zoom in, it becomes a line” intuition works. A corner never becomes a single line no matter how much you zoom; it keeps its sharpness.

A quick comparison table #

Feature at x=aContinuous?Derivative exists?What you see
Smooth curveYesYesOne clear tangent slope
Corner (x)Yes
Vertical tangent (∛x)YesNo (finite)Slopes blow up toward ±∞
Jump discontinuityNoNoBreak in graph

What the canvas should show in this section #

To make “failure modes” memorable:

Application/Connection: What Derivatives Enable #

Derivatives as a sensitivity tool #

At a point aaa, the derivative tells you how sensitive the output is to tiny input changes.

If Δx\Delta xΔx is small, then

Δf≈f′(a) Δx\Delta f \approx f'(a),\Delta xΔf≈f′(a)Δx

This is the bridge from geometry to practical estimation.

Example intuition: If a cost function has derivative 5 dollars/unit at the current production level, then producing one more unit increases cost by about $5.

Derivatives as slope fields for behavior #

Even before learning derivative “rules,” knowing that f′(x)f'(x)f′(x) is slope gives a powerful way to interpret a graph:

This connects directly to optimization ideas used everywhere (including machine learning).

Why this node unlocks the next nodes #

The definition is conceptually perfect but computationally slow. The next node (Derivative Rules) gives shortcuts (power/product/quotient/chain rules). Those rules are justified because they match what the limit definition produces.

In MLE, you maximize a likelihood (or log-likelihood) with respect to parameters. “Maximize” often means “take derivative, set it to zero, solve.” The derivative is what turns “best parameter” into an equation you can solve.

Convexity uses derivatives to formalize “curves that bend upward.” For differentiable functions, one hallmark is that the derivative is increasing. So understanding f′(x)f'(x)f′(x) as slope makes convexity feel natural.

Integrals and derivatives are paired by the Fundamental Theorem of Calculus. Informally: differentiation measures instantaneous change, integration accumulates change. This node provides the “change” half of that story.

A small, concrete bridge: tangent line approximation #

Once you can compute f′(a)f'(a)f′(a), you can approximate fff near aaa by its tangent line:

f(x)≈f(a)+f′(a)(x−a)f(x) \approx f(a) + f'(a)(x-a)f(x)≈f(a)+f′(a)(x−a)

This idea appears constantly later: numerical methods, error estimates, optimization steps, and more.

Worked Examples (3) #

Compute a derivative from the definition: f(x)=x² at an arbitrary point a #

Find f′(a)f'(a)f′(a) for f(x)=x2f(x)=x^2f(x)=x2 using the defining limit.

Goal: turn the limit into algebra and simplify until the limit is easy.

  1. Start with the definition:

    f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}f′(a)=h→0lim​hf(a+h)−f(a)​

  2. Plug in f(x)=x2f(x)=x^2f(x)=x2:

    f′(a)=lim⁡h→0(a+h)2−a2hf'(a)=\lim_{h\to 0}\frac{(a+h)^2-a^2}{h}f′(a)=h→0lim​h(a+h)2−a2​

  3. Expand (a+h)2(a+h)^2(a+h)2:

    (a+h)2=a2+2ah+h2(a+h)^2=a^2+2ah+h^2(a+h)2=a2+2ah+h2

    So the numerator becomes:

    (a2+2ah+h2)−a2=2ah+h2(a^2+2ah+h^2)-a^2=2ah+h^2(a2+2ah+h2)−a2=2ah+h2

  4. Factor out hhh:

    2ah+h2h=h(2a+h)h=2a+h\frac{2ah+h^2}{h}=\frac{h(2a+h)}{h}=2a+hh2ah+h2​=hh(2a+h)​=2a+h

    (For h≠0h\neq 0h=0 this cancellation is valid; the limit cares about values near 0, not at 0.)

  5. Now take the limit:

    f′(a)=lim⁡h→0(2a+h)=2af'(a)=\lim_{h\to 0}(2a+h)=2af′(a)=h→0lim​(2a+h)=2a

Insight: The limit definition often works by algebraically canceling the problematic hhh in the denominator. For polynomials like x2x^2x2, the derivative emerges cleanly: f′(a)=2af'(a)=2af′(a)=2a.

Non-differentiability at a corner: f(x)=|x| at x=0 #

Show that f′(0)f'(0)f′(0) does not exist for f(x)=∣x∣f(x)=|x|f(x)=∣x∣.

Strategy: compute the right-hand and left-hand derivatives and compare.

  1. Start with the difference quotient at a=0a=0a=0:

    f(0+h)−f(0)h=∣h∣−0h=∣h∣h\frac{f(0+h)-f(0)}{h}=\frac{|h|-0}{h}=\frac{|h|}{h}hf(0+h)−f(0)​=h∣h∣−0​=h∣h∣​

  2. Right-hand derivative (approach with h>0h>0h>0):

    If h>0h>0h>0, then ∣h∣=h|h|=h∣h∣=h, so

    ∣h∣h=hh=1\frac{|h|}{h}=\frac{h}{h}=1h∣h∣​=hh​=1

    Thus

    f+′(0)=lim⁡h→0+1=1f'_+(0)=\lim_{h\to 0^+} 1 = 1f+′​(0)=h→0+lim​1=1

  3. Left-hand derivative (approach with h<0h<0h<0):

    If h<0h<0h<0, then ∣h∣=−h|h|=-h∣h∣=−h, so

    ∣h∣h=−hh=−1\frac{|h|}{h}=\frac{-h}{h}=-1h∣h∣​=h−h​=−1

    Thus

    f−′(0)=lim⁡h→0−(−1)=−1f'_-(0)=\lim_{h\to 0^-} (-1) = -1f−′​(0)=h→0−lim​(−1)=−1

  4. Compare:

    f+′(0)=1≠−1=f−′(0)f'_+(0)=1 \neq -1 = f'_-(0)f+′​(0)=1=−1=f−′​(0)

    Therefore, the two-sided limit does not exist, so f′(0)f'(0)f′(0) does not exist.

Insight: A corner is exactly the situation where the curve has two competing tangent directions. The derivative requires a single slope; disagreement between one-sided slopes means ‘not differentiable.’

Tangent line from the derivative: f(x)=x² at x=3 #

Use the derivative to find the equation of the tangent line to y=x2y=x^2y=x2 at x=3x=3x=3.

You can use the result from Example 1: if f(x)=x2f(x)=x^2f(x)=x2, then f′(x)=2xf'(x)=2xf′(x)=2x (so f′(3)=6f'(3)=6f′(3)=6).

  1. Compute the point on the curve:

    f(3)=32=9f(3)=3^2=9f(3)=32=9

    So the point is (3,9)(3,9)(3,9).

  2. Compute the slope of the tangent line:

    f′(x)=2x⇒f′(3)=6f'(x)=2x \Rightarrow f'(3)=6f′(x)=2x⇒f′(3)=6

  3. Use point-slope form:

    y−9=6(x−3)y-9=6(x-3)y−9=6(x−3)

  4. Simplify (optional):

    y−9=6x−18⇒y=6x−9y-9=6x-18 \Rightarrow y=6x-9y−9=6x−18⇒y=6x−9

Insight: Once you know the derivative at a point, geometry becomes straightforward: slope + point gives the tangent line immediately.

Key Takeaways #

Common Mistakes #

Practice #

easy

Use the definition to compute the derivative of f(x)=3x+2f(x)=3x+2f(x)=3x+2 at an arbitrary point aaa.

Hint: Compute f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h}hf(a+h)−f(a)​, simplify, then take h→0h\to 0h→0.

Show solution

Start:

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}f′(a)=h→0lim​hf(a+h)−f(a)​

Compute:

f(a+h)=3(a+h)+2=3a+3h+2f(a+h)=3(a+h)+2=3a+3h+2f(a+h)=3(a+h)+2=3a+3h+2

f(a)=3a+2f(a)=3a+2f(a)=3a+2

Difference quotient:

(3a+3h+2)−(3a+2)h=3hh=3\frac{(3a+3h+2)-(3a+2)}{h}=\frac{3h}{h}=3h(3a+3h+2)−(3a+2)​=h3h​=3

Limit:

f′(a)=lim⁡h→03=3f'(a)=\lim_{h\to 0}3=3f′(a)=h→0lim​3=3

medium

Find the slope of the tangent line to f(x)=1xf(x)=\frac{1}{x}f(x)=x1​ at x=2x=2x=2 using the definition of the derivative.

Hint: Compute 12+h−12h\frac{\frac{1}{2+h}-\frac{1}{2}}{h}h2+h1​−21​​, combine fractions, simplify, then take h→0h\to 0h→0.

Show solution

Definition at a=2a=2a=2:

f′(2)=lim⁡h→012+h−12hf'(2)=\lim_{h\to 0}\frac{\frac{1}{2+h}-\frac{1}{2}}{h}f′(2)=h→0lim​h2+h1​−21​​

Combine the numerator:

12+h−12=2−(2+h)2(2+h)=−h2(2+h)\frac{1}{2+h}-\frac{1}{2}=\frac{2-(2+h)}{2(2+h)}=\frac{-h}{2(2+h)}2+h1​−21​=2(2+h)2−(2+h)​=2(2+h)−h​

Divide by hhh:

−h2(2+h)h=−h2(2+h)h=−12(2+h)\frac{\frac{-h}{2(2+h)}}{h}=\frac{-h}{2(2+h)h}=-\frac{1}{2(2+h)}h2(2+h)−h​​=2(2+h)h−h​=−2(2+h)1​

Take the limit:

f′(2)=lim⁡h→0(−12(2+h))=−12⋅2=−14f'(2)=\lim_{h\to 0}\left(-\frac{1}{2(2+h)}\right)=-\frac{1}{2\cdot 2}=-\frac{1}{4}f′(2)=h→0lim​(−2(2+h)1​)=−2⋅21​=−41​

medium

Determine whether f(x)=∣x−1∣f(x)=|x-1|f(x)=∣x−1∣ is differentiable at x=1x=1x=1. Justify using one-sided derivatives (you may use the definition conceptually without heavy algebra).

Hint: Shift the |x| corner: ∣x−1∣|x-1|∣x−1∣ has the same shape as ∣x∣|x|∣x∣ but centered at 1. Compare the slope from the left vs right.

Show solution

At x=1x=1x=1, the graph of ∣x−1∣|x-1|∣x−1∣ has a corner (it’s ∣x∣|x|∣x∣ shifted right by 1). For x>1x>1x>1, ∣x−1∣=x−1|x-1|=x-1∣x−1∣=x−1 which has slope 1. For x<1x<1x<1, ∣x−1∣=−(x−1)=1−x|x-1|=-(x-1)=1-x∣x−1∣=−(x−1)=1−x which has slope −1. Since the right-hand derivative is 1 and the left-hand derivative is −1, they do not match. Therefore fff is not differentiable at x=1x=1x=1.

Connections #

Quality: A (4.1/5)

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