In short

The derivative of a function f(x) is the limit f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. In physics, this turns a position function x(t) into velocity v = \frac{dx}{dt}, and velocity into acceleration a = \frac{dv}{dt}. The derivative gives you the instantaneous rate of change — the exact value at one moment, not an average over an interval.

An autorickshaw speeds through a narrow lane in Jaipur. The meter reads 28 km/h. A moment later it reads 35 km/h. Somewhere between those two readings, the driver hit the accelerator. But when exactly did the speed change? And how fast was the autorickshaw moving at the precise instant the driver's foot pressed the pedal — not a second before, not a second after, but at that single frozen moment?

The speedometer answers this question in practice. Calculus answers it in theory. The tool is called the derivative, and it is the single most important mathematical idea in all of physics. Every time a physicist writes "the velocity at time t," they mean a derivative. Every time you see "the rate at which temperature changes," that is a derivative. Acceleration, current, power, pressure gradient — all derivatives. Physics without differentiation is like cricket without a bat.

The problem: averages are not enough

Suppose you drop a coconut from the top of a 45 m tall coconut tree. Its height above the ground, in metres, after t seconds is:

h(t) = 45 - 4.9t^2

Why this formula: the coconut starts at 45 m and falls freely under gravity. The term 4.9t^2 is \frac{1}{2}g t^2 with g = 9.8 m/s².

How fast is the coconut moving at t = 2 s?

You might try computing the average speed between t = 2 and t = 3:

\text{average speed} = \frac{h(2) - h(3)}{3 - 2} = \frac{(45 - 19.6) - (45 - 44.1)}{1} = \frac{25.4 - 0.9}{1} = 24.5 \text{ m/s}

But that is the average over a full second. The coconut is accelerating, so it moves faster at t = 3 than at t = 2. The average overshates the speed at t = 2.

Try a smaller interval — from t = 2 to t = 2.1:

\frac{h(2) - h(2.1)}{0.1} = \frac{25.4 - (45 - 21.609)}{0.1} = \frac{25.4 - 23.391}{0.1} = \frac{2.009}{0.1} = 20.09 \text{ m/s}

Smaller still — t = 2 to t = 2.01:

\frac{h(2) - h(2.01)}{0.01} = \frac{25.4 - (45 - 19.796)}{0.01} = \frac{25.4 - 25.204}{0.01} = \frac{0.196}{0.01} = 19.6 \text{ m/s}
Interval Average speed (m/s)
t = 2 to t = 3 24.5
t = 2 to t = 2.1 20.09
t = 2 to t = 2.01 19.60
t = 2 to t = 2.001 19.60

The numbers are converging to 19.6 m/s. As the interval shrinks toward zero, the average speed locks onto the instantaneous speed at t = 2 s. The derivative is the mathematical machine that performs this shrinking exactly.

The derivative — from first principles

Definition of the derivative

The derivative of a function f(x) at a point x is:

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

when this limit exists.

Reading the definition. Look at each piece. f(x) is the value of the function at the point you care about. f(x+h) is the value a tiny step h away. The numerator f(x+h) - f(x) is how much the function changed over that step. Dividing by h turns the change into a rate — change per unit of x. And \lim_{h \to 0} says: shrink the step until it vanishes. What survives is the rate at the single point x.

Geometrically, this is the slope of the tangent line to the curve at the point (x, f(x)). When h is finite, the ratio \frac{f(x+h) - f(x)}{h} is the slope of a secant line connecting two points on the curve. As h \to 0, the secant rotates into the tangent — and its slope becomes the derivative.

Interactive: secant line approaching the tangent line on a parabola A parabola y = 5x squared. A fixed point at x = 2 and a draggable point move along the curve. A secant line connects them. As the draggable point approaches the fixed point, the secant becomes the tangent. Readouts show the slope. x f(x) = 5x² 1 2 3 drag the red point
Drag the red point along the curve. The tangent line's slope updates live — this slope is the derivative $f'(x) = 10x$ at whatever point you pick. At $x = 2$, the slope is 20. At $x = 3$, the slope is 30.

Deriving the derivative of h(t) = 45 - 4.9t^2 at t = 2

Apply the definition directly.

Step 1. Compute h(2 + h).

h(2 + h) = 45 - 4.9(2 + h)^2 = 45 - 4.9(4 + 4h + h^2) = 45 - 19.6 - 19.6h - 4.9h^2
h(2 + h) = 25.4 - 19.6h - 4.9h^2

Why: expand (2 + h)^2 using the binomial formula. Multiply each term by 4.9.

Step 2. Compute the difference h(2 + h) - h(2).

h(2) = 45 - 19.6 = 25.4
h(2 + h) - h(2) = (25.4 - 19.6h - 4.9h^2) - 25.4 = -19.6h - 4.9h^2

Why: the constant terms cancel. What remains are the terms that depend on h — the parts that measure how much h(t) changed.

Step 3. Divide by h.

\frac{h(2 + h) - h(2)}{h} = \frac{-19.6h - 4.9h^2}{h} = -19.6 - 4.9h

Why: every term in the numerator has a factor of h, so dividing by h is clean — no 0/0 problems yet.

Step 4. Take the limit as h \to 0.

\lim_{h \to 0} (-19.6 - 4.9h) = -19.6

Why: as h shrinks to zero, the -4.9h term vanishes. The rate that survives is -19.6 m/s. The negative sign means the height is decreasing — the coconut is falling.

\boxed{h'(2) = -19.6 \text{ m/s}}

The coconut's instantaneous speed at t = 2 s is 19.6 m/s downward. This matches exactly what the table of averages was converging to — the derivative has done the infinite shrinking in one clean algebraic step.

Velocity and acceleration — the physics of derivatives

The derivative's physical meaning is the instantaneous rate of change. In mechanics, the two most important derivatives are:

Velocity: the derivative of position

If a body's position is x(t), its velocity is:

v(t) = \frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{x(t + \Delta t) - x(t)}{\Delta t}

Why: velocity is how fast position changes. The derivative gives the exact rate at each instant, not an average over a journey.

Acceleration: the derivative of velocity

If the velocity is v(t), the acceleration is:

a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2}

Why: acceleration is how fast velocity changes. It is the derivative of velocity, which makes it the second derivative of position — the rate of change of the rate of change.

This chain — position \xrightarrow{\frac{d}{dt}} velocity \xrightarrow{\frac{d}{dt}} acceleration — is the backbone of mechanics. When you know the position function x(t), one differentiation gives you the velocity and a second gives you the acceleration. The ISRO engineers tracking Chandrayaan's trajectory do exactly this: they have a position function in three dimensions, and they differentiate it to get the spacecraft's velocity and acceleration at every instant.

Derivative rules — the toolkit

Computing derivatives from the limit definition every time is possible but slow. These rules are shortcuts, each derivable from first principles.

The power rule

\frac{d}{dx}(x^n) = nx^{n-1}

Why: for any real number n, bring the exponent down as a multiplier and reduce the exponent by 1. This covers polynomials, square roots (n = \frac{1}{2}), and reciprocals (n = -1).

Derivation (for positive integers). Start with f(x) = x^n. By the definition:

f'(x) = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}

Factor (x+h)^n - x^n using the identity a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1}), with a = x+h, b = x:

(x+h)^n - x^n = h\bigl[(x+h)^{n-1} + (x+h)^{n-2}x + \cdots + x^{n-1}\bigr]

Why: this factoring pulls out the h = (x+h) - x that will cancel with the denominator.

Divide by h:

\frac{(x+h)^n - x^n}{h} = (x+h)^{n-1} + (x+h)^{n-2}x + \cdots + x^{n-1}

Take \lim_{h \to 0}: every term becomes x^{n-1}, and there are n such terms:

f'(x) = nx^{n-1}

Trigonometric derivatives

Function Derivative Physics use
\sin x \cos x SHM displacement \to velocity
\cos x -\sin x Velocity \to acceleration in SHM
\tan x \sec^2 x Angle of incline problems

Why: the derivative of \sin x is \cos x because the rate at which the sine changes is largest when the sine itself is zero (at the peaks and troughs of the cosine). In simple harmonic motion, the displacement is x = A\sin(\omega t) and the velocity is v = A\omega\cos(\omega t) — the velocity leads the displacement by a quarter cycle.

Exponential and logarithmic derivatives

Function Derivative Physics use
e^x e^x Radioactive decay, capacitor discharge
a^x a^x \ln a Population growth models
\ln x \frac{1}{x} Entropy, Boltzmann factor

Why: e^x is the unique function that is its own derivative — its rate of growth at any point equals its current value. This is exactly why exponential functions describe processes where the rate of change is proportional to the current amount, like nuclear decay: the more radioactive nuclei you have, the faster they decay.

Sum and constant-multiple rules

\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)
\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)

Why: differentiation is a linear operation. You can differentiate a sum term by term, and constants slide out. This means you can differentiate any polynomial by applying the power rule to each term.

The chain rule — derivatives of compositions

A rocket launched by ISRO burns fuel at a rate that depends on altitude, and the altitude depends on time. To find how the fuel burn rate changes with time, you need to connect two rates: fuel-vs-altitude and altitude-vs-time. The chain rule is the connector.

If y depends on u and u depends on x, then:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Why: the rate at which y changes with respect to x is the rate at which y changes with u, multiplied by the rate at which u changes with x. The intermediate variable u links the two.

Physical meaning. Suppose the temperature T of a metal rod depends on position x along the rod, and a heat wave is moving along the rod so that the position of peak temperature changes with time: x = x(t). Then the rate at which temperature changes at the wave front is:

\frac{dT}{dt} = \frac{dT}{dx} \cdot \frac{dx}{dt}

The first factor is the temperature gradient (how sharply temperature varies along the rod). The second factor is the speed of the wave front. Their product gives the temporal rate of temperature change at the moving front.

Example. A cricket ball's height is h = 20t - 5t^2, and the air pressure varies with height as P(h) = P_0 \, e^{-h/8500} (a simplified barometric formula). Find \frac{dP}{dt} at t = 1 s.

Step 1. Find \frac{dh}{dt}.

\frac{dh}{dt} = 20 - 10t

At t = 1: \frac{dh}{dt} = 10 m/s.

Why: differentiate using the power rule on each term. The ball is rising at 10 m/s at t = 1 s.

Step 2. Find \frac{dP}{dh}.

\frac{dP}{dh} = P_0 \cdot \left(-\frac{1}{8500}\right) e^{-h/8500}

At t = 1: h = 20 - 5 = 15 m, so \frac{dP}{dh} = -\frac{P_0}{8500}\,e^{-15/8500} \approx -\frac{P_0}{8500} (since e^{-0.00176} \approx 1).

Why: differentiate the exponential using the chain rule itself — the derivative of e^{f(h)} is e^{f(h)} \cdot f'(h).

Step 3. Apply the chain rule.

\frac{dP}{dt} = \frac{dP}{dh} \cdot \frac{dh}{dt} \approx -\frac{P_0}{8500} \times 10 = -\frac{P_0}{850} \text{ Pa/s}

Why: the pressure drops as the ball rises. The rate of pressure change combines two effects — how sharply pressure falls with height, and how fast the ball is climbing.

Maxima and minima — finding extremes

When a physical quantity reaches its maximum or minimum, its rate of change is momentarily zero — it is neither increasing nor decreasing. This is the key to optimization in physics: set the derivative equal to zero and solve.

A curve showing a maximum where the tangent is horizontal A smooth curve rises to a peak and falls. At the peak, a horizontal tangent line is drawn. An arrow points to the peak, labeled "dy/dx = 0 here". dy/dx = 0 x y
At a maximum, the tangent line is horizontal — the derivative is zero. The function switches from increasing (positive slope) to decreasing (negative slope).

The procedure

  1. Write the quantity as a function of one variable.
  2. Differentiate and set \frac{dy}{dx} = 0 — solve for x.
  3. Check the second derivative \frac{d^2y}{dx^2} at that point:
    • If \frac{d^2y}{dx^2} < 0, the point is a maximum (the curve is concave down).
    • If \frac{d^2y}{dx^2} > 0, the point is a minimum (the curve is concave up).

Why the second derivative test works: the second derivative tells you how the slope itself is changing. At a maximum, the slope is going from positive to negative — it is decreasing — so its derivative (the second derivative) is negative.

Where physics needs this

When physics requires differentiation — slopes and sensitivity

Beyond velocity and acceleration, differentiation appears everywhere in physics:

Slope of a tangent line. Any time you read a physical quantity off a graph, and that quantity changes, the slope of the curve at that point is a derivative. The slope of a displacement–time graph is velocity. The slope of a velocity–time graph is acceleration. The slope of a charge–time graph is current. The slope of a potential energy–distance graph, with a minus sign, is force: F = -\frac{dU}{dx}.

Sensitivity analysis. Suppose a quantity Q depends on a measurement x. How sensitive is Q to small errors in x? The answer is the derivative \frac{dQ}{dx}. If you measure the radius of a steel sphere to calculate its volume V = \frac{4}{3}\pi r^3, a small error \delta r in the radius produces an error:

\delta V \approx \frac{dV}{dr} \cdot \delta r = 4\pi r^2 \cdot \delta r

Why: the derivative converts a small change in the input (\delta r) to the resulting change in the output (\delta V). For a sphere, the volume is more sensitive to radius errors when r is large — because 4\pi r^2 grows with r.

Worked examples

Example 1: Finding velocity and acceleration from a position function

An autorickshaw moves along a straight road. Its position in metres at time t seconds is:

x(t) = 3t^2 - 2t + 1

Find the velocity and acceleration at t = 2 s.

Step 1. Differentiate x(t) to get velocity.

v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 2t + 1)

Apply the power rule to each term:

v(t) = 3 \cdot 2t^{2-1} - 2 \cdot 1t^{1-1} + 0 = 6t - 2

Why: the derivative of 3t^2 is 6t (bring down the 2, reduce the power by 1). The derivative of -2t is -2 (a linear term differentiates to its coefficient). The derivative of a constant (1) is 0 — a constant does not change.

Step 2. Evaluate at t = 2 s.

v(2) = 6(2) - 2 = 12 - 2 = 10 \text{ m/s}

Why: substitute t = 2 directly into the velocity function. The autorickshaw is moving at 10 m/s at this instant.

Step 3. Differentiate v(t) to get acceleration.

a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t - 2) = 6 \text{ m/s}^2

Why: the derivative of 6t is 6. The derivative of -2 is 0. The acceleration is constant — this autorickshaw accelerates at a steady 6 m/s² regardless of time.

Step 4. Verify the result makes physical sense.

At t = 0: x(0) = 1 m, v(0) = -2 m/s (moving backward briefly), a = 6 m/s² (constant).

At t = 2: x(2) = 12 - 4 + 1 = 9 m, v(2) = 10 m/s (now moving forward and fast).

Why: the negative initial velocity means the autorickshaw was briefly reversing at t = 0. The constant positive acceleration of 6 m/s² quickly reverses this and pushes it forward. By t = 2, it has covered 9 m from the origin and is moving at 10 m/s.

Solid curve: position $x(t) = 3t^2 - 2t + 1$, a parabola opening upward. Dashed line: velocity $v(t) = 6t - 2$, a straight line. At $t = 2$, the autorickshaw is at 9 m with a velocity of 10 m/s.

Result: At t = 2 s, the velocity is v = 10 m/s and the acceleration is a = 6 m/s².

What this shows: One differentiation takes you from position to velocity. A second differentiation takes you from velocity to acceleration. The position curve is a parabola, its derivative is a line, and the derivative of that line is a constant — each differentiation reduces the degree of the polynomial by 1.

Example 2: Maximum height of a projectile

A cricket ball is hit straight up with an initial speed of 20 m/s. Its height above the ground is:

h(t) = 20t - 5t^2

Find the maximum height and the time at which the ball reaches it.

Step 1. Differentiate to find the velocity.

v(t) = \frac{dh}{dt} = 20 - 10t

Why: the derivative of 20t is 20, and the derivative of -5t^2 is -10t. This velocity starts at 20 m/s and decreases by 10 m/s every second — the ball is decelerating under gravity.

Step 2. Set v(t) = 0 to find the time of maximum height.

20 - 10t = 0 \implies t = 2 \text{ s}

Why: at the peak, the ball momentarily stops moving — its velocity is zero. This is the instant before it starts falling back down.

Step 3. Verify this is a maximum using the second derivative.

a(t) = \frac{d^2h}{dt^2} = -10 \text{ m/s}^2

Since \frac{d^2h}{dt^2} = -10 < 0, the curve is concave down, confirming this is a maximum.

Why: the second derivative is the acceleration due to gravity (-g \approx -10 m/s²). Gravity always pulls downward, so the height curve always bends downward — every stationary point is a maximum. There are no minima for a projectile in free flight (ignoring the ground).

Step 4. Compute the maximum height.

h(2) = 20(2) - 5(2)^2 = 40 - 20 = 20 \text{ m}
The cricket ball's height $h(t) = 20t - 5t^2$. The trajectory is a downward-opening parabola. The peak at $t = 2$ s, $h = 20$ m is exactly where the derivative (velocity) equals zero.

Result: The ball reaches a maximum height of 20 m at t = 2 s.

What this shows: Setting the derivative to zero finds the exact moment when the ball is at its highest. The second derivative (acceleration) confirms this is a maximum, not a minimum. Every projectile problem about "maximum height" or "maximum range" is a derivative-equals-zero problem.

Common confusions

If you came here to understand what a derivative is, how to compute one, and how physics uses derivatives for velocity, acceleration, and optimization, you have what you need. What follows is for readers who want the formal underpinning and some advanced applications.

The formal \varepsilon-\delta definition

The limit \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = L means: for every \varepsilon > 0, there exists a \delta > 0 such that

0 < |h| < \delta \implies \left|\frac{f(x+h) - f(x)}{h} - L\right| < \varepsilon

This is the rigorous version of "as h gets close to 0, the difference quotient gets close to L." The \varepsilon measures how close you want the quotient to be to L, and the \delta is the guarantee that making h small enough achieves that closeness.

For f(x) = x^2, verifying f'(x) = 2x:

\left|\frac{(x+h)^2 - x^2}{h} - 2x\right| = |2x + h - 2x| = |h| < \delta

Choose \delta = \varepsilon. Done. The proof is clean because the difference quotient simplifies to 2x + h, and the error is exactly |h|.

Product and quotient rules

The product rule: \frac{d}{dx}[f(x)\,g(x)] = f'(x)\,g(x) + f(x)\,g'(x)

Derivation. Start from the definition:

\frac{d}{dx}[fg] = \lim_{h \to 0}\frac{f(x+h)\,g(x+h) - f(x)\,g(x)}{h}

Add and subtract f(x+h)\,g(x) in the numerator:

= \lim_{h \to 0}\frac{f(x+h)\,g(x+h) - f(x+h)\,g(x) + f(x+h)\,g(x) - f(x)\,g(x)}{h}
= \lim_{h \to 0}\left[f(x+h)\cdot\frac{g(x+h) - g(x)}{h} + g(x)\cdot\frac{f(x+h) - f(x)}{h}\right]
= f(x)\,g'(x) + g(x)\,f'(x)

Why: the trick is adding and subtracting f(x+h)\,g(x). This creates two difference quotients — one for g and one for f — that each converge to their respective derivatives.

Physics application: Power delivered to a body. If force F(t) and velocity v(t) both vary with time, the power is P = Fv, and the rate of change of power is:

\frac{dP}{dt} = \frac{dF}{dt}\,v + F\,\frac{dv}{dt} = \dot{F}\,v + F\,a

The quotient rule: \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{[g(x)]^2}

Differentiability and continuity

A function can be continuous at a point without being differentiable there. The classic example is f(x) = |x| at x = 0: the function is continuous (no gap), but the graph has a sharp corner, and the slope jumps from -1 to +1 at the origin. There is no single tangent line, so the derivative does not exist.

In physics, such points matter. A ball bouncing off a hard floor has a velocity-time graph with a sharp corner at the moment of bounce — the velocity jumps from -v to +v instantaneously (in the idealised model). At that instant, acceleration is undefined (or, more precisely, infinite — a Dirac delta impulse). Real collisions smooth out this corner over a very short contact time, making the acceleration large but finite.

Higher-order derivatives and jerk

The third derivative of position, \frac{d^3x}{dt^3}, is called jerk. It measures how suddenly the acceleration changes. Jerk is what you feel when a Delhi Metro train starts or stops — not the acceleration itself, but the onset of acceleration. A smooth ride means low jerk.

Amusement park rides are engineered to control jerk carefully. Too much jerk and riders feel whiplash. The position function of a well-designed ride transition is a polynomial of degree 5 or higher, because you need enough degrees of freedom to control position, velocity, acceleration, and jerk simultaneously at the start and end of the transition.

Where this leads next