What a derivative is
A derivative measures how fast one quantity changes as another changes. If $y$ is a function of $x$, written $y = f(x)$, the derivative $\dfrac{dy}{dx}$ tells you the rate at which $y$ responds to a change in $x$ at a particular point. The same symbol does the work of every "rate" in physics: position against time gives velocity, velocity against time gives acceleration, charge against time gives current. Learn to read $\dfrac{dy}{dx}$ as "the rate of change of $y$ with respect to $x$" and the physics translates itself.
Geometrically, the derivative is the slope of the tangent to the curve $y = f(x)$ at the chosen point. A steep curve has a large derivative; a flat curve has a derivative near zero; a curve sloping downward has a negative derivative. This geometric picture is the key that unlocks graph questions: the velocity at an instant is the steepness of the position–time graph right at that instant, not the height of the graph.
The limit definition (conceptual)
To pin down "instantaneous", start with the average rate over a finite interval. If $x$ increases by $\Delta x$ and $y$ responds by $\Delta y$, the average rate of change is $\dfrac{\Delta y}{\Delta x}$ — the slope of the secant line joining the two points. This is an average over the whole interval; it does not tell you the rate at a single instant.
The derivative is what the average rate approaches as the interval shrinks to nothing:
$$\frac{dy}{dx} \;=\; \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} \;=\; \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}.$$
As $\Delta x \to 0$ the secant pivots into the tangent, and the average rate becomes the instantaneous rate. NEET does not ask you to evaluate this limit by hand — but you must understand it, because it is exactly why velocity is the slope of the $x$–$t$ graph at an instant rather than the average velocity over an interval. The limit converts "average over a stretch" into "rate at a point".
Standard derivatives — the table
In NEET you do not differentiate from first principles. You memorise a short table and combine its entries with the four rules. Commit the following to memory; nearly every derivative you will ever take in physics is built from these seven results.
| Function $y$ | Derivative $dy/dx$ | Note |
|---|---|---|
| $x^{n}$ | $n\,x^{n-1}$ | Power rule, valid for any real $n$. $\tfrac{d}{dx}(x^3)=3x^2$; $\tfrac{d}{dx}(\sqrt{x})=\tfrac{1}{2\sqrt{x}}$. |
| constant $c$ | $0$ | A constant has zero slope. This is the mirror of the integration constant $C$. |
| $\sin x$ | $\cos x$ | $x$ in radians. Underlies SHM, AC, waves. |
| $\cos x$ | $-\sin x$ | The minus sign is the SHM restoring-force sign — do not lose it. |
| $\tan x$ | $\sec^2 x$ | Diverges at $x=\pi/2$. |
| $e^{x}$ | $e^{x}$ | The function that is its own derivative; drives exponential growth and decay. |
| $\ln x$ | $\dfrac{1}{x}$ | Natural log (base $e$). For $\log_{10}$, multiply by $0.4343$. |
Two utility results follow at once from the power rule and the constant rule: the derivative of a straight line $y = mx + c$ is simply $m$ (constant slope), and the derivative of $x$ itself is $1$. A constant multiplier rides along untouched: $\tfrac{d}{dx}(k\,x^n) = k\,n\,x^{n-1}$.
Differentiation has an exact inverse. Once you can read this table backwards you can reverse rates into totals — see Integration — basics for physics.
The four rules of differentiation
The table gives you the derivatives of single, "bare" functions. The four rules below let you differentiate any combination of them. Three of the four are mechanical; the fourth — the chain rule — is the one NEET keeps catching students on.
| Rule | Statement | Example |
|---|---|---|
| Sum rule | $\dfrac{d}{dx}(u+v) = \dfrac{du}{dx} + \dfrac{dv}{dx}$ | $\tfrac{d}{dx}(x^2 + \sin x) = 2x + \cos x$ |
| Product rule | $\dfrac{d}{dx}(uv) = u'v + uv'$ | $\tfrac{d}{dx}(x^2\sin x) = 2x\sin x + x^2\cos x$ |
| Quotient rule | $\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^2}$ | $\tfrac{d}{dx}\!\left(\tfrac{\sin x}{x}\right) = \tfrac{x\cos x - \sin x}{x^2}$ |
| Chain rule | $\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}$ | $\tfrac{d}{dt}\sin(\omega t) = \omega\cos(\omega t)$ |
The chain rule is the workhorse. If $y$ depends on $u$ and $u$ depends on $x$, you differentiate the outer function and multiply by the derivative of the inside. Almost every physics derivative is secretly a chain-rule derivative: $\tfrac{d}{dt}\sin(\omega t) = \omega\cos(\omega t)$ — the inside $\omega t$ contributes its derivative $\omega$ out front; $\tfrac{d}{dx}(2x+1)^5 = 5(2x+1)^4 \cdot 2$; $\tfrac{d}{dt}\,e^{-\lambda t} = -\lambda\,e^{-\lambda t}$ — the decay constant $-\lambda$ is pulled out. Whenever anything other than a bare variable sits inside a function, the chain rule is waiting.
The second derivative
Differentiate once and you get a rate. Differentiate the result again and you get the rate at which that rate is changing — the second derivative, written $\dfrac{d^2y}{dx^2}$. In kinematics this is the chain that links position, velocity and acceleration: velocity is the first derivative of position, $v = \dfrac{dx}{dt}$, and acceleration is the derivative of velocity, so
$$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}.$$
Acceleration is therefore the second derivative of position with respect to time. The sign of the second derivative also reports the concavity of a curve: a positive second derivative means the curve bends upward (concave up, like a valley), a negative one means it bends downward (concave down, like the peak of a projectile's path). That sign is exactly what distinguishes a minimum from a maximum, which is the next section.
Maxima and minima
At the top of a hill or the bottom of a valley, a smooth curve is momentarily flat — its tangent is horizontal, so its slope is zero. This gives the first-derivative test for locating extreme points: set $\dfrac{dy}{dx} = 0$ and solve. Each solution is a stationary point, a candidate maximum or minimum. To classify it, look at the second derivative:
| Condition at the stationary point | Type of point | Shape |
|---|---|---|
| $\dfrac{dy}{dx}=0$ and $\dfrac{d^2y}{dx^2} < 0$ | Maximum | Concave down — peak (e.g. top of a projectile path) |
| $\dfrac{dy}{dx}=0$ and $\dfrac{d^2y}{dx^2} > 0$ | Minimum | Concave up — trough |
| $\dfrac{dy}{dx}=0$ and $\dfrac{d^2y}{dx^2}=0$ | Inconclusive | Inspect the sign change of $dy/dx$ around the point |
This is the engine behind a whole family of NEET problems: the maximum height of a projectile (where the vertical velocity $\tfrac{dy}{dt}=0$), the range that is largest at $45^\circ$, the time of maximum current in a circuit, the angle that minimises the time of a journey. The recipe never changes — differentiate, set to zero, solve, then confirm the type with the second derivative.
The physics meaning — v, a and graph slopes
Differentiation is not an abstract exercise in NEET physics; it is the definition of velocity and acceleration. Position is a function of time, $x(t)$. Its first derivative is the instantaneous velocity, and the derivative of velocity is the instantaneous acceleration:
$$v(t) = \frac{dx}{dt}, \qquad a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}.$$
Read on a graph, these become statements about slope. The velocity at any instant is the slope of the tangent to the position–time ($x$–$t$) curve at that instant. The acceleration at any instant is the slope of the tangent to the velocity–time ($v$–$t$) curve. A straight, upward-sloping $x$–$t$ line means constant velocity; a curving $x$–$t$ graph means the velocity is changing, i.e. there is acceleration. On a $v$–$t$ graph a horizontal line means zero acceleration, and a sloping line means constant acceleration equal to that slope.
Worked examples from kinematics
Each example is the same routine: write the position or velocity function, apply the table and the rules, and read off the physics. Nothing here is a novel calculation — it is the table applied with care.
A particle moves along a line with position $x(t) = 4t^2 - 3t + 7$ (SI units). Find its velocity and acceleration, and the instant at which it is momentarily at rest.
Velocity. Differentiate term by term using the power rule and constant rule: $v = \dfrac{dx}{dt} = 8t - 3$. The constant $7$ has zero derivative; the $-3t$ becomes $-3$; the $4t^2$ becomes $8t$.
Acceleration. Differentiate again: $a = \dfrac{dv}{dt} = 8\ \text{m s}^{-2}$ — constant. Equivalently $a = \dfrac{d^2x}{dt^2} = 8$.
Momentarily at rest. Set $v = 0$: $8t - 3 = 0 \Rightarrow t = 0.375\ \text{s}$. At that instant the slope of the $x$–$t$ graph is zero, even though $x$ is non-zero there.
A projectile launched vertically has height $y(t) = u t - \tfrac{1}{2}g t^2$. Use differentiation to find the time of maximum height and the maximum height, with $u = 20\ \text{m s}^{-1}$ and $g = 10\ \text{m s}^{-2}$.
Vertical velocity. $\dfrac{dy}{dt} = u - g t$. At maximum height the slope of the $y$–$t$ graph is zero, so $u - g t = 0 \Rightarrow t = \dfrac{u}{g} = \dfrac{20}{10} = 2\ \text{s}$.
Confirm it is a maximum. $\dfrac{d^2y}{dt^2} = -g = -10 < 0$, so the stationary point is indeed a maximum (the path is concave down).
Maximum height. Substitute $t = 2\ \text{s}$ into $y(t)$: $y = 20(2) - \tfrac{1}{2}(10)(2)^2 = 40 - 20 = 20\ \text{m}$. Note that at the top the velocity is zero, not the height.
A body oscillates with displacement $x(t) = A\sin(\omega t)$. Find its velocity and acceleration, and show that the acceleration is proportional to $-x$.
Velocity (chain rule). The inside is $\omega t$, whose derivative is $\omega$, so $v = \dfrac{dx}{dt} = A\omega\cos(\omega t)$.
Acceleration (chain rule again). Differentiate $\cos(\omega t)$ to get $-\omega\sin(\omega t)$ and carry the inner $\omega$: $a = \dfrac{dv}{dt} = -A\omega^2\sin(\omega t)$.
Read the physics. Since $A\sin(\omega t) = x$, the acceleration is $a = -\omega^2 x$ — proportional to displacement and oppositely directed. That is the defining equation of simple harmonic motion, obtained purely by differentiating twice. Dropping the inner $\omega$ in either step would destroy the $\omega^2$ and the SHM relation.
Where this shows up in NEET physics
Mathematical Tools is never graded directly — it is tested through every mechanics, oscillations and electromagnetism question. These are the recurring places differentiation does the work.
Velocity and acceleration from a position function.
Given any $x(t)$, the velocity is $v = dx/dt$ and the acceleration is $a = dv/dt = d^2x/dt^2$. Questions that give a polynomial or trigonometric $x(t)$ and ask for $v$ or $a$ at a stated time, or for the instant of rest ($v=0$), reduce to one or two applications of the power and chain rules.
Slope of $x$–$t$ and $v$–$t$ graphs.
The velocity is the slope of the $x$–$t$ graph; the acceleration is the slope of the $v$–$t$ graph. Graph-matching questions hinge on reading steepness, not height: a flat $x$–$t$ portion means $v=0$, a straight sloping $v$–$t$ line means constant acceleration. See Motion in a Straight Line.
Maximum height, optimum angle, peak current.
Setting $dy/dx = 0$ locates the extreme value; the sign of $d^2y/dx^2$ classifies it. This solves projectile maximum height (vertical velocity zero), the $45^\circ$ angle for maximum range, and the time at which a current or charge peaks in a circuit.
SHM and exponential rates via the chain rule.
Differentiating $x = A\sin(\omega t)$ twice gives $a = -\omega^2 x$, the SHM signature. Differentiating $q = q_0 e^{-t/\tau}$ gives the current $i = dq/dt = -(q_0/\tau)e^{-t/\tau}$. Both rely on carrying the inner derivative ($\omega$ or $-1/\tau$) correctly.
FAQs — Differentiation for Physics
Short answers to the differentiation questions NEET aspirants stumble on most.
Do I need the limit definition of the derivative for NEET?
What is the difference between the slope of an x–t graph and its value?
Why must angles be in radians when differentiating sin x and cos x?
How does the chain rule apply to sin(ωt) and e^(−λt)?
How do I tell a maximum from a minimum using derivatives?
Is acceleration the second derivative of position?
What is the derivative of a constant, and why does it matter in physics?
Differentiation in one breath
- The derivative $dy/dx$ is the instantaneous rate of change and the slope of the tangent — the limit of $\Delta y/\Delta x$ as $\Delta x \to 0$.
- Memorise seven derivatives: $x^n \to n x^{n-1}$, constant $\to 0$, $\sin x \to \cos x$, $\cos x \to -\sin x$, $\tan x \to \sec^2 x$, $e^x \to e^x$, $\ln x \to 1/x$.
- Four rules: sum, product $u'v+uv'$, quotient $(u'v-uv')/v^2$, and the chain rule $\tfrac{dy}{dx}=\tfrac{dy}{du}\tfrac{du}{dx}$ — always carry the inner derivative.
- Velocity $v = dx/dt$; acceleration $a = dv/dt = d^2x/dt^2$. Velocity is the slope of the $x$–$t$ graph; acceleration is the slope of the $v$–$t$ graph.
- Maxima/minima: set $dy/dx = 0$, then $d^2y/dx^2 < 0$ for a maximum, $> 0$ for a minimum.
- Slope ≠ value: zero slope marks a peak or rest, not zero position.