# Applications of Differentiation

Mathematics I · Differential Calculus · https://tryals.app/learn/mathematics-i/applications-of-differentiation

## The Reach of the Derivative

Wherever something changes, the **derivative** describes how fast. The unifying insight is that derivatives convert static formulas into dynamic rates, forming the language of quantitative science.

| Application | Formula / Meaning | Description |
| --- | --- | --- |
| **Motion** | $x'(t)$ = **velocity**, $x''(t)$ = **acceleration** | Kinematics tracks position; $F = ma$ links force to the second derivative |
| **Growth/Decay** | $\frac{dN}{dt} = kN$ | **Exponential growth** ($k>0$) and decay ($k<0$); solution $N = N_0 e^{kt}$ |

**Common Pitfall:** Viewing these applications as separate topics. They are the same **derivative** applied to different quantities: velocity is the rate of position, and marginal cost is the rate of total cost.

## Optimization & Approximations

Setting a derivative to zero finds the best choice — minimum cost, maximum profit, or shortest time. From economics to engineering, optimization is ubiquitous.

| Application | Concept & Formula |
| --- | --- |
| **Optimization** | Setting $f'(x) = 0$ finds optimal values; marginal cost in economics is a derivative |
| **Approximation** | **Newton's method** uses tangent lines to find roots; **differentials** estimate errors |

Mastering differentiation means asking of any changing system: what is its rate, and what does that rate imply?

## Practice questions

7 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Marginal cost at a quantity q is the total cost at q divided by q.

**Answer:** False

**Why:** False. That formula is average cost; marginal cost is the DERIVATIVE of total cost. Fixed costs inflate the average but vanish from the derivative, which is exactly why the two diverge and why decisions about producing one more unit use the marginal figure.

Page: https://tryals.app/practice/mathematics-i/applications-of-differentiation/marginal-cost-at-a-quantity-q-is-the-total-cost-at-q-divided-by-q

### 2. Velocity, acceleration and marginal cost are all derivatives of quite different things. What do the three have in common?

A. Each measures a total accumulated over an interval
B. Each is an instantaneous rate of change of one quantity per another
C. Each requires the underlying function to be a polynomial
D. Each is the slope of a straight line through two points

**Answer:** B. Each is an instantaneous rate of change of one quantity per another

**Why:** One operation, many readings. Marginal cost is $dC/dq$ exactly as velocity is $dx/dt$, which is why an economist and a physicist use the same calculus.

Page: https://tryals.app/practice/mathematics-i/applications-of-differentiation/velocity-acceleration-and-marginal-cost-are-all-derivatives-of-quite

### 3. Velocity, marginal cost and population growth rate are all derivatives. Why does one mathematical operation describe such unrelated quantities?

A. Because these three quantities happen to be governed by similar equations, which is a coincidence of their fields.
B. Because the derivative is defined separately in each field and the shared name is historical.
C. Because physics, economics and biology were unified by calculus into a single discipline.
D. Because all three ask how fast one thing changes as another does.

**Answer:** D. Because all three ask how fast one thing changes as another does.

**Why:** Write the three out and one question remains once the nouns are stripped: how fast does this change as that changes. It is the same abstraction argument as the vector-space lesson, prove it once and every field gets its results free. Marginal cost is not an economic idea borrowed from physics; it is the same derivative asked about a different pair of variables.

Page: https://tryals.app/practice/mathematics-i/applications-of-differentiation/velocity-marginal-cost-and-population-growth-rate-are-all

### 4. Sort each quantity by whether it is itself the derivative of some other quantity.

**Answer:**

- Is a derivative of something else: Velocity, Acceleration, Marginal cost, The growth rate of a population
- Is not a derivative: Total cost, Position

**Why:** Velocity is the rate of position, acceleration the rate of velocity, marginal cost the rate of total cost, and the growth rate the rate of population. Position and total cost are the underlying totals, each application pairs a total with its rate, and the derivative is always the bridge.

Page: https://tryals.app/practice/mathematics-i/applications-of-differentiation/sort-each-quantity-by-whether-it-is-itself-the-derivative-of-some

### 5. Newton's method usually converges spectacularly fast and sometimes flies off entirely. What single feature of the step explains both behaviours?

A. The division by f'(x), which throws the guess far if the slope is flat
B. The use of a tangent rather than a secant, which is always less stable
C. The number of iterations, since too many cause divergence
D. The starting point, which determines the outcome independently of the function

**Answer:** A. The division by f'(x), which throws the guess far if the slope is flat

**Why:** One term does both jobs: a steep tangent points almost exactly at the root and the correct digits roughly double each step, while a flat one divides by nearly zero and launches the iterate somewhere useless. That is why production solvers keep a bracketing method underneath, speed when it works, a guarantee when it does not.

Page: https://tryals.app/practice/mathematics-i/applications-of-differentiation/newtons-method-usually-converges-spectacularly-fast-and-sometimes

### 6. The model $N = N_{0}e^{kt}$ solves dN/dt = kN. Select every TRUE statement.

A. A negative k gives decay
B. N_0 is the amount present at time zero
C. The rate dN/dt is proportional to the current amount N
D. The rate dN/dt stays constant over time
E. A positive k gives growth

**Answer:** A. A negative k gives decay; B. N_0 is the amount present at time zero; C. The rate dN/dt is proportional to the current amount N; E. A positive k gives growth

**Why:** The defining equation makes the rate PROPORTIONAL to the current amount, so as N grows the rate grows too, that is what distinguishes exponential growth from linear growth, where the rate would be constant. Setting t = 0 gives N = N_0, identifying the initial amount.

Page: https://tryals.app/practice/mathematics-i/applications-of-differentiation/the-model-n-n0e-kt-solves-dn-dt-kn-select-every-true-statement

### 7. L'Hôpital's rule replaces the limit of f/g by the limit of f'/g' - but only for indeterminate forms like 0/0. Why does applying it to an ordinary limit give the wrong answer?

A. Because differentiating a quotient requires the quotient rule, which the rule omits.
B. Because the rule is not a general identity about quotients.
C. Because f'/g' is always larger than f/g, so the rule overestimates non-indeterminate limits.
D. Because the rule requires the limit to exist beforehand, which is unknown for ordinary limits.

**Answer:** B. Because the rule is not a general identity about quotients.

**Why:** Test it where the form is not indeterminate: (x+1)/(x+2) at 0 is 1/2 by substitution, and differentiating top and bottom gives 1. The proof runs through the Cauchy Mean Value Theorem and depends on both functions vanishing, so check the form every time, including after each application. It can also fail to help: x/√(x²+1) returns you to a similar expression indefinitely.

Page: https://tryals.app/practice/mathematics-i/applications-of-differentiation/lhopitals-rule-replaces-the-limit-of-f-g-by-the-limit-of-f-g