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Mathematical Tools & Measurement

  1. 01Basic Essential Mathematics
  2. 02Vectors
  3. 03Differentiation
  4. 04Applications of Differentiation
  5. 05Integration
  6. 06Applications of Integration
  7. 07Physical Quantities and Units
  8. 08Dimensional Formula
  9. 09Dimensional Analysis and Its Applications
  10. 10Experimental Skills

Kinematics

  1. 01Motion in One Dimension
  2. 02Motion in Multiple Dimensions
  3. 03Relative Velocity
  4. 04Circular Motion Kinematics
  5. 05Circular Motion Dynamics

Dynamics

  1. 01Forces and Laws of Motion
  2. 02Laws of Motion
  3. 03Friction
  4. 04Force and Potential Energy
  5. 05Fundamentals of Force
  6. 06Newton's Laws and Free Body Diagrams
  7. 07Applications: Objects in Equilibrium
  8. 08Applications: Objects in Motion
  9. 09Constraint Relations
  10. 10Inertial and Non-Inertial Frames
  11. 11Basics of Friction
  12. 12Applications of Friction

Work, Energy, and Power

  1. 01Conservation of Mechanical Energy
  2. 02Work and the Work-Energy Theorem
  3. 03Work and Kinetic Energy Theorem
  4. 04Energy and its Conservation
Theory/Mathematical Tools & Measurement

Mathematical Tools & Measurement · Chapter 03

Differentiation

Differentiation detailed theory study guide for Physics.

68 min read · 7 topics

01

The Concept of the Derivative

From Average to Instantaneous Change

Consider a road trip. Your average speed for the whole trip might be 60 km/h. But at any single moment, your speedometer might read 100 km/h or 0 km/h. The 60 km/h is an average rate of change over a long time interval. The speedometer reading is the instantaneous rate of change. How do we calculate that 'at a moment' value? This is the central question the derivative answers.

The derivative is the mathematical tool for finding the instantaneous rate of change of a function.

The Geometric View: Slope of a Tangent

We can see this on a graph. The average rate of change between two points on a curve is simply the slope of the straight line connecting them (a secant line).

Average Slope=ΔyΔx=f(x+Δx)−f(x)Δx\text{Average Slope} = \frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x}Average Slope=ΔxΔy​=Δxf(x+Δx)−f(x)​

To find the instantaneous rate of change at a single point, we need to make the interval Δx\Delta xΔx incredibly small. Slide the second point closer and closer to the first point. The secant line will pivot, getting closer and closer to becoming the tangent line—the line that just touches the curve at that single point.

The secant through P and Q and the tangent at P drawn on one curve, with the run Δx and the rise Δy of the secant marked.
The secant through P and Q and the tangent at P drawn on one curve, with the run Δx and the rise Δy of the secant marked.

The sequence below shows this pivoting more explicitly for a point P0(x,y)P_0(x,y)P0​(x,y): as the second point P1P_1P1​ is brought closer along the curve, the secant P0P1P_0P_1P0​P1​ rotates until it settles into the tangent, inclined at the final angle α\alphaα to the x-axis.

The secant through P and Q tilting, panel by panel, into the tangent at P, which makes the final angle α with the horizontal
The secant through P and Q tilting, panel by panel, into the tangent at P, which makes the final angle α with the horizontal

The slope of this tangent line is the instantaneous rate of change at that point. The derivative *is* the slope of the tangent line.

This also gives a quick way to read the behaviour of a function directly off its derivative: if the function is increasing with xxx, its derivative is positive; if the function is decreasing, its derivative is negative; and if the function is momentarily unchanged, its derivative is zero.

The Formal Definition of the Derivative

To capture this idea of making the interval 'infinitely small', we use the concept of a limit. The derivative of a function f(x)f(x)f(x) with respect to xxx is defined as:

dydx=f′(x)=lim⁡Δx→0f(x+Δx)−f(x)Δx\frac{dy}{dx} = f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}dxdy​=f′(x)=Δx→0lim​Δxf(x+Δx)−f(x)​

This fundamental formula is the basis of differential calculus. It takes the expression for the slope of a secant line and finds its limit as the interval size shrinks to zero.

Common Notations

  • Leibniz Notation: dydx\frac{dy}{dx}dxdy​ - emphasizes the 'change in y' over 'change in x'.
  • Lagrange Notation: f′(x)f'(x)f′(x) (read 'f prime of x') - compact and function-focused.
  • Newton's Notation: x˙\dot{x}x˙ - commonly used in physics specifically for derivatives with respect to time (e.g., velocity).

The Derivative in Physics

The derivative is arguably the most important mathematical tool in physics. It allows us to define many fundamental quantities:

  • Velocity: The instantaneous rate of change of position with respect to time.
    v=dxdtv = \frac{dx}{dt}v=dtdx​
  • Acceleration: The instantaneous rate of change of velocity with respect to time.
    a=dvdta = \frac{dv}{dt}a=dtdv​
  • Force: The rate of change of momentum with respect to time.
    F=dpdtF = \frac{dp}{dt}F=dtdp​
  • Current: The rate of flow of charge with respect to time.
    I=dQdtI = \frac{dQ}{dt}I=dtdQ​

The slope of the tangent line to a position-time graph at a specific time 't' represents what physical quantity?

02

Key Concepts

Key concepts will be added here.

03

Rules of Differentiation

The Differentiation Toolkit

Finding the derivative from the limit definition every time is tedious. Mathematicians have developed a set of powerful, time-saving rules. Mastering these rules is the key to becoming proficient in calculus.

Standard Derivatives Reference Table

Before applying the rules below, it helps to have the derivatives of the most common functions on hand. Each of these can itself be derived from the limit definition, but in practice they are used as a ready reference.

Function yyyDerivative dydx\dfrac{dy}{dx}dxdy​
Any constant000
xnx^nxnnxn−1nx^{n-1}nxn−1
sin⁡x\sin xsinxcos⁡x\cos xcosx
cos⁡x\cos xcosx−sin⁡x-\sin x−sinx
tan⁡x\tan xtanxsec⁡2x\sec^2 xsec2x
cot⁡x\cot xcotx−cosec2x-\text{cosec}^2 x−cosec2x
sec⁡x\sec xsecxsec⁡xtan⁡x\sec x \tan xsecxtanx
cosec x\text{cosec}\,xcosecx−cosec xcot⁡x-\text{cosec}\,x\cot x−cosecxcotx
exe^xexexe^xex
axa^xaxaxln⁡aa^x \ln aaxlna
ln⁡x\ln xlnx1x\dfrac{1}{x}x1​
log⁡ax\log_a xloga​x1xln⁡a\dfrac{1}{x\ln a}xlna1​
sin⁡−1x\sin^{-1} xsin−1x11−x2,  ∣x∣<1\dfrac{1}{\sqrt{1-x^2}}, \; |x|<11−x2​1​,∣x∣<1
cos⁡−1x\cos^{-1} xcos−1x−11−x2,  ∣x∣<1\dfrac{-1}{\sqrt{1-x^2}}, \; |x|<11−x2​−1​,∣x∣<1
tan⁡−1x\tan^{-1} xtan−1x11+x2\dfrac{1}{1+x^2}1+x21​
cot⁡−1x\cot^{-1} xcot−1x−11+x2\dfrac{-1}{1+x^2}1+x2−1​
sec⁡−1x\sec^{-1} xsec−1x1xx2−1,  ∣x∣>1\dfrac{1}{x\sqrt{x^2-1}}, \; |x|>1xx2−1​1​,∣x∣>1
cosec−1x\text{cosec}^{-1} xcosec−1x−1xx2−1,  ∣x∣>1\dfrac{-1}{x\sqrt{x^2-1}}, \; |x|>1xx2−1​−1​,∣x∣>1

1. The Power Rule

This is the most common rule you will use. It applies to any term of the form xnx^nxn.

ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}dxd​(xn)=nxn−1

In words: Bring the power down in front as a multiplier, then subtract one from the power.

  • Example 1: ddx(x5)=5x5−1=5x4\frac{d}{dx}(x^5) = 5x^{5-1} = 5x^4dxd​(x5)=5x5−1=5x4
  • Example 2 (Square Root): ddx(x)=ddx(x1/2)=12x−1/2=12x\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}dxd​(x​)=dxd​(x1/2)=21​x−1/2=2x​1​

(Also, the derivative of a constant is 0, and the derivative of a constant times a function is the constant times the derivative of the function.)

2. The Product Rule

Used when you have two functions of x multiplied together, like f(x)=u(x)v(x)f(x) = u(x)v(x)f(x)=u(x)v(x).

ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}dxd​(uv)=udxdv​+vdxdu​

In words: The first function times the derivative of the second, plus the second function times the derivative of the first.

Example: Differentiate y=x2sin⁡(x)y = x^2 \sin(x)y=x2sin(x).
Here, u=x2u=x^2u=x2 and v=sin⁡(x)v=\sin(x)v=sin(x).
dydx=x2ddx(sin⁡(x))+sin⁡(x)ddx(x2)\frac{dy}{dx} = x^2 \frac{d}{dx}(\sin(x)) + \sin(x) \frac{d}{dx}(x^2)dxdy​=x2dxd​(sin(x))+sin(x)dxd​(x2)
dydx=x2cos⁡(x)+2xsin⁡(x)\frac{dy}{dx} = x^2\cos(x) + 2x\sin(x)dxdy​=x2cos(x)+2xsin(x)

3. The Quotient Rule

Used when you have one function divided by another, f(x)=u(x)/v(x)f(x) = u(x)/v(x)f(x)=u(x)/v(x).

ddx(uv)=vdudx−udvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}dxd​(vu​)=v2vdxdu​−udxdv​​

In words: 'Low D-high minus High D-low, over the square of what's below.'

Example: Differentiate y=x3x2+1y = \frac{x^3}{x^2+1}y=x2+1x3​.
Here, u=x3u=x^3u=x3 and v=x2+1v=x^2+1v=x2+1.
dydx=(x2+1)(3x2)−(x3)(2x)(x2+1)2=3x4+3x2−2x4(x2+1)2=x4+3x2(x2+1)2\frac{dy}{dx} = \frac{(x^2+1)(3x^2) - (x^3)(2x)}{(x^2+1)^2} = \frac{3x^4+3x^2 - 2x^4}{(x^2+1)^2} = \frac{x^4+3x^2}{(x^2+1)^2}dxdy​=(x2+1)2(x2+1)(3x2)−(x3)(2x)​=(x2+1)23x4+3x2−2x4​=(x2+1)2x4+3x2​

4. The Chain Rule

This rule is used for 'nested' functions (a function inside of a function). If y=f(g(x))y = f(g(x))y=f(g(x)), we let u=g(x)u = g(x)u=g(x) so y=f(u)y = f(u)y=f(u).

dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​

In words: The derivative of the outer function (with the inside function left alone) times the derivative of the inner function.

Example: Differentiate y=(x2+1)3y = (x^2+1)^3y=(x2+1)3.
The outer function is u3u^3u3 and the inner function is u=x2+1u=x^2+1u=x2+1.
dydx=(3(x2+1)2)⋅(2x)=6x(x2+1)2\frac{dy}{dx} = (3(x^2+1)^2) \cdot (2x) = 6x(x^2+1)^2dxdy​=(3(x2+1)2)⋅(2x)=6x(x2+1)2

Using the chain rule, what is the derivative of y=cos⁡(x2)y = \cos(x^2)y=cos(x2)?

04

Examples

Putting the Limit Definition to Work

The formal limit definition of the derivative can feel abstract until you use it directly on a real quantity. The following examples build the idea of instantaneous velocity from the ground up, starting with an average rate of change over a finite interval and then shrinking that interval to zero.

Example 17: Average Velocity from a Position Function

The position xxx (in metres) of a particle depends on time ttt (in seconds) as x=t2x = t^2x=t2. The average velocity of a particle is defined as the total displacement in a time interval divided by that time interval. Find the average velocity of the particle (i) in the first two seconds, and (ii) from t=3 st=3\,\text{s}t=3s to t=5 st=5\,\text{s}t=5s.

Solution:

(i) Average velocity in the first two seconds:

vˉ=Δx in 2s2s=(2)2−(0)2(2)−(0)=4 m2 s=2 m/s\bar{v} = \frac{\Delta x \text{ in } 2\text{s}}{2\text{s}} = \frac{(2)^2 - (0)^2}{(2) - (0)} = \frac{4\,\text{m}}{2\,\text{s}} = 2\,\text{m/s}vˉ=2sΔx in 2s​=(2)−(0)(2)2−(0)2​=2s4m​=2m/s

(ii) Average velocity from 3 s to 5 s:

vˉ=Δx from 3s to 5s2s=(5)2−(3)2(5)−(3)=16 m2 s=8 m/s\bar{v} = \frac{\Delta x \text{ from 3s to 5s}}{2\text{s}} = \frac{(5)^2 - (3)^2}{(5) - (3)} = \frac{16\,\text{m}}{2\,\text{s}} = 8\,\text{m/s}vˉ=2sΔx from 3s to 5s​=(5)−(3)(5)2−(3)2​=2s16m​=8m/s

Example 18: Instantaneous Velocity via the Limit Definition

Consider the same particle, x=t2x = t^2x=t2. The instantaneous velocity is defined as the average velocity as the time interval tends to zero. Find the instantaneous velocity of the particle (i) at t=2 st=2\,\text{s}t=2s, (ii) at t=5 st=5\,\text{s}t=5s, and (iii) at any general time ttt.

Solution:

(i) The instantaneous velocity at 2 s is the rate of change of xxx with respect to ttt at t=2 st=2\,\text{s}t=2s. We consider a small time interval Δt\Delta tΔt after 2 s and take the limit of the ratio of the displacement in this interval to Δt\Delta tΔt:

v(2)=lim⁡Δt→0ΔxΔt=lim⁡Δt→0(2+Δt)2−(2)2(2+Δt)−(2)v(2) = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \lim_{\Delta t \to 0} \frac{(2+\Delta t)^2 - (2)^2}{(2+\Delta t) - (2)}v(2)=Δt→0lim​ΔtΔx​=Δt→0lim​(2+Δt)−(2)(2+Δt)2−(2)2​

=lim⁡Δt→0(4+4Δt+Δt2)−4Δt=lim⁡Δt→0Δt(4+Δt)Δt= \lim_{\Delta t \to 0} \frac{(4 + 4\Delta t + \Delta t^2) - 4}{\Delta t} = \lim_{\Delta t \to 0} \frac{\Delta t(4+\Delta t)}{\Delta t}=Δt→0lim​Δt(4+4Δt+Δt2)−4​=Δt→0lim​ΔtΔt(4+Δt)​

Since Δt\Delta tΔt is very small but not equal to zero, it can be cancelled from numerator and denominator, giving

v(2)=lim⁡Δt→0(4+Δt)=4 m/sv(2) = \lim_{\Delta t \to 0}(4+\Delta t) = 4\,\text{m/s}v(2)=Δt→0lim​(4+Δt)=4m/s

(ii) Repeating the same process for a small interval Δt\Delta tΔt after 5 s:

v(5)=lim⁡Δt→0(5+Δt)2−(5)2(5+Δt)−(5)=lim⁡Δt→0(25+10Δt+Δt2)−25Δt=lim⁡Δt→0Δt(10+Δt)Δtv(5) = \lim_{\Delta t \to 0} \frac{(5+\Delta t)^2 - (5)^2}{(5+\Delta t) - (5)} = \lim_{\Delta t \to 0} \frac{(25 + 10\Delta t + \Delta t^2) - 25}{\Delta t} = \lim_{\Delta t \to 0} \frac{\Delta t(10+\Delta t)}{\Delta t}v(5)=Δt→0lim​(5+Δt)−(5)(5+Δt)2−(5)2​=Δt→0lim​Δt(25+10Δt+Δt2)−25​=Δt→0lim​ΔtΔt(10+Δt)​

v(5)=lim⁡Δt→0(10+Δt)=10 m/sv(5) = \lim_{\Delta t \to 0}(10+\Delta t) = 10\,\text{m/s}v(5)=Δt→0lim​(10+Δt)=10m/s

(iii) Finally, for a general time ttt, the same steps give

v(t)=lim⁡Δt→0(t+Δt)2−t2(t+Δt)−t=lim⁡Δt→0(t2+2tΔt+Δt2)−t2Δt=lim⁡Δt→0Δt(2t+Δt)Δtv(t) = \lim_{\Delta t \to 0} \frac{(t+\Delta t)^2 - t^2}{(t+\Delta t) - t} = \lim_{\Delta t \to 0} \frac{(t^2 + 2t\Delta t + \Delta t^2) - t^2}{\Delta t} = \lim_{\Delta t \to 0} \frac{\Delta t(2t+\Delta t)}{\Delta t}v(t)=Δt→0lim​(t+Δt)−t(t+Δt)2−t2​=Δt→0lim​Δt(t2+2tΔt+Δt2)−t2​=Δt→0lim​ΔtΔt(2t+Δt)​

v(t)=lim⁡Δt→0(2t+Δt)=2tv(t) = \lim_{\Delta t \to 0}(2t+\Delta t) = 2tv(t)=Δt→0lim​(2t+Δt)=2t

Notice that this general result v(t)=2tv(t) = 2tv(t)=2t reproduces both earlier answers: v(2)=2(2)=4 m/sv(2)=2(2)=4\,\text{m/s}v(2)=2(2)=4m/s and v(5)=2(5)=10 m/sv(5)=2(5)=10\,\text{m/s}v(5)=2(5)=10m/s. This is exactly the power rule in action — differentiating x=t2x=t^2x=t2 directly from the limit definition gives dxdt=2t\dfrac{dx}{dt}=2tdtdx​=2t.

Applying the Standard Rules

Once the derivative table and the sum, product, and chain rules are known, most derivatives can be found without ever returning to the limit definition. The following examples show these rules in action.

Example 19: Power Rule

Determine the derivative of the function y=x5y = x^5y=x5.

Solution: Using the standard-derivatives table with n=5n=5n=5:

d(x5)dx=5x5−1=5x4\frac{d(x^5)}{dx} = 5x^{5-1} = 5x^4dxd(x5)​=5x5−1=5x4

Example 20: Sum Rule

Determine the derivative of the function y=x3+x2y = x^3 + x^2y=x3+x2.

Solution: Using the standard-derivatives table, d(x3)dx=3x2\dfrac{d(x^3)}{dx} = 3x^2dxd(x3)​=3x2 and d(x2)dx=2x\dfrac{d(x^2)}{dx} = 2xdxd(x2)​=2x. Applying the sum rule,

d(x3+x2)dx=3x2+2x\frac{d(x^3+x^2)}{dx} = 3x^2 + 2xdxd(x3+x2)​=3x2+2x

Example 21: Constant Multiple and Sum Rules Together

Determine the derivative of the function y=2x12+3x23y = 2x^{12} + 3x^{23}y=2x12+3x23.

Solution: Using the power rule together with the constant-multiple rule, d(2x12)dx=2×12x11=24x11\dfrac{d(2x^{12})}{dx} = 2\times12x^{11} = 24x^{11}dxd(2x12)​=2×12x11=24x11 and d(3x23)dx=3×23x22=69x22\dfrac{d(3x^{23})}{dx} = 3\times23x^{22} = 69x^{22}dxd(3x23)​=3×23x22=69x22. Applying the sum rule,

d(2x12+3x23)dx=24x11+69x22\frac{d(2x^{12}+3x^{23})}{dx} = 24x^{11} + 69x^{22}dxd(2x12+3x23)​=24x11+69x22

Example 22: Product Rule

Find the derivative of y=xln⁡xy = x\ln xy=xlnx with respect to xxx.

Solution: Using the Product Rule with u=xu=xu=x and v=ln⁡xv=\ln xv=lnx,

dydx=xddx(ln⁡x)+(ln⁡x)ddx(x)\frac{dy}{dx} = x\frac{d}{dx}(\ln x) + (\ln x)\frac{d}{dx}(x)dxdy​=xdxd​(lnx)+(lnx)dxd​(x)

dydx=x(1x)+ln⁡x=1+ln⁡x\frac{dy}{dx} = x\left(\frac{1}{x}\right) + \ln x = 1 + \ln xdxdy​=x(x1​)+lnx=1+lnx

Example 23: Chain Rule

Find the derivative of y=sin⁡(x2)y = \sin(x^2)y=sin(x2) with respect to xxx.

Solution: This requires the Chain Rule. Let z=x2z = x^2z=x2, so that y=sin⁡zy = \sin zy=sinz. Then

dydx=dydz⋅dzdx=ddz(sin⁡z)⋅ddx(x2)=(cos⁡z)(2x)=2xcos⁡z\frac{dy}{dx} = \frac{dy}{dz}\cdot\frac{dz}{dx} = \frac{d}{dz}(\sin z)\cdot\frac{d}{dx}(x^2) = (\cos z)(2x) = 2x\cos zdxdy​=dzdy​⋅dxdz​=dzd​(sinz)⋅dxd​(x2)=(cosz)(2x)=2xcosz

Since z=x2z = x^2z=x2, this gives

dydx=2xcos⁡(x2)\frac{dy}{dx} = 2x\cos(x^2)dxdy​=2xcos(x2)

A particle's position is x=t2x = t^2x=t2 (x in metres, t in seconds). What is its instantaneous velocity at t=6 st = 6\ \text{s}t=6 s?

What is ddx(xln⁡x)\dfrac{d}{dx}(x\ln x)dxd​(xlnx)?

05

Higher Order Derivatives

Differentiating a Derivative

Since the derivative of a function, f′(x)f'(x)f′(x), is itself a function, we can take its derivative as well. This is called the second derivative, and we can continue this process to find the third, fourth, and even higher-order derivatives.

Notation

  • Second Derivative: f′′(x)f''(x)f′′(x) or d2ydx2\frac{d^2y}{dx^2}dx2d2y​
  • Third Derivative: f′′′(x)f'''(x)f′′′(x) or d3ydx3\frac{d^3y}{dx^3}dx3d3y​

The Meaning of the Second Derivative

The second derivative tells us the rate of change of the rate of change. This might sound abstract, but it has two very important and intuitive interpretations.

1. Physical Meaning: Acceleration

In physics, the second derivative is most famously known as acceleration.

  • Position, x(t)x(t)x(t): Where an object is.
  • First Derivative, v(t)=dxdtv(t) = \frac{dx}{dt}v(t)=dtdx​: The rate of change of position (how fast it's moving).
  • Second Derivative, a(t)=dvdt=d2xdt2a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}a(t)=dtdv​=dt2d2x​: The rate of change of velocity (how fast its speed or direction is changing).

A positive acceleration means the velocity is increasing, while a negative acceleration (deceleration) means the velocity is decreasing.

2. Geometric Meaning: Concavity

Geometrically, the second derivative tells us about the concavity of a function's graph—which way the curve is bending.

  • If f′′(x)>0f''(x) > 0f′′(x)>0, the graph is concave up (like a cup holding water). The slope is increasing.
  • If f′′(x)<0f''(x) < 0f′′(x)<0, the graph is concave down (like a cup spilling water). The slope is decreasing.

Concave up, where f''(x) &gt; 0 and the tangent slopes increase; concave down, where f''(x) &lt; 0 and they decrease.
Concave up, where f''(x) > 0 and the tangent slopes increase; concave down, where f''(x) < 0 and they decrease.

Example: Position, Velocity, and Acceleration

Suppose an object's position is given by the function x(t)=t3−6t2+9t+1x(t) = t^3 - 6t^2 + 9t + 1x(t)=t3−6t2+9t+1.

1. Find the velocity, v(t)v(t)v(t):
v(t)=dxdt=3t2−12t+9v(t) = \frac{dx}{dt} = 3t^2 - 12t + 9v(t)=dtdx​=3t2−12t+9

2. Find the acceleration, a(t)a(t)a(t):
a(t)=dvdt=d2xdt2=6t−12a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2} = 6t - 12a(t)=dtdv​=dt2d2x​=6t−12

Using these, we can find the velocity and acceleration at any time. For instance, at t=1t=1t=1, the object has zero velocity (v(1)=0v(1)=0v(1)=0) but a negative acceleration (a(1)=−6a(1)=-6a(1)=−6), meaning it is momentarily at rest but about to move in the negative direction.

If a car is moving at a constant velocity, what must be true about its position function, x(t)?

06

Introduction to Differentiation

Welcome to Differentiation. Content to be added.

07

Limits and Continuity: The Foundation of Calculus

The Need for Limits

Calculus is the mathematics of change. But to understand change at a single instant, we first need a tool to describe what happens as we get infinitely close to that instant. This tool is the limit. The concept of a limit is the foundation on which all of calculus is built.

The Idea of a Limit

A limit answers the question: 'What value is a function approaching as its input approaches a certain point?'

Consider the function f(x)=x2f(x) = x^2f(x)=x2. What value does this function 'approach' as xxx gets closer and closer to 2? It's easy to see that as xxx gets closer to 2, f(x)f(x)f(x) gets closer to 4. We write this formally as:

lim⁡x→2x2=4\lim_{x \to 2} x^2 = 4x→2lim​x2=4

This reads: 'The limit of x2x^2x2 as xxx approaches 2 is 4.'

Why Not Just Plug In the Number?

In the example above, we could have just plugged in x=2x=2x=2 to get 4. But consider this function:

g(x)=x2−4x−2g(x) = \frac{x^2 - 4}{x - 2}g(x)=x−2x2−4​

What is the limit as xxx approaches 2? If we plug in x=2x=2x=2, we get 00\frac{0}{0}00​, which is undefined. But a limit doesn't care about what happens *at* the point, only what happens as we get *infinitely close* to it. For any x≠2x \neq 2x=2, we can simplify the function: g(x)=(x−2)(x+2)x−2=x+2g(x) = \frac{(x-2)(x+2)}{x-2} = x+2g(x)=x−2(x−2)(x+2)​=x+2. Now, as xxx gets closer and closer to 2, g(x)g(x)g(x) gets closer and closer to 4. So, lim⁡x→2g(x)=4\lim_{x \to 2} g(x) = 4limx→2​g(x)=4, even though the function itself is undefined at that point.

The graph of g(x) has an open circle at x = 2, yet the arrows closing in from both sides approach the value 4.
The graph of g(x) has an open circle at x = 2, yet the arrows closing in from both sides approach the value 4.

Continuity: An Unbroken Path

Now that we understand limits, we can define continuity. A function is continuous at a point if it meets three conditions:

  1. The function is defined at that point (the point exists).
  2. The limit of the function exists at that point (the function approaches a single value from both sides).
  3. The limit and the function's value are the same.

Intuitively, a continuous function is one whose graph you can draw without lifting your pen from the paper. There are no sudden jumps, holes, or breaks.

Left: a continuous curve with no breaks. Right: a jump discontinuity, where the two branches fail to meet.
Left: a continuous curve with no breaks. Right: a jump discontinuity, where the two branches fail to meet.

Most functions we deal with in introductory physics (like polynomials) are continuous everywhere, which makes our analysis easier. The concept of continuity is what allows us to assume that change happens smoothly.

A function is defined as f(x) = 1/x. This function is discontinuous at which point?

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