Understanding Differentiation: A Simple Guide to Key Formulas

Master differentiation in minutes! Learn all key formulas and rules in one simple guide — perfect for students and exam preparation

By :  ravi
Update: 2025-10-27 12:34 GMT

Mathematics seems to be tough for students, especially those who are preparing for competitive exams like IIT JEE. These subjects like Physics, Chemistry, and Maths, demand hard work, practice, patience, and persistence.

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One of the core concepts in Maths is Differentiation. It is important to know the critical equations and ideas in order to solve problems. Understanding the fundamentals, students can tackle challenging problems with ease, being time efficient while not making mistakes typical for young learners.

We feel that the prep for IIT JEE should not start very late. We at MNR Scottsdale start to cook the potential right from class VIII and above so that they grow as complete beings, developing their mother wit, skill of analysis and problem-solving approach. We enable students to realize their dreams of getting into IITs by strengthening their fundamentals and making them practice constantly.



Understanding Differentiation: A Simple Guide to Key Formulas

Differentiation is one of the most important concepts in calculus. It helps measure how a function changes with respect to its variable — for example, how distance changes over time. The process of finding the derivative of a function is called differentiation, and it has applications in physics, engineering, economics, and data science. Let’s explore the main formulas and rules in simple terms.

Differentiation and integration are two fundamental operations in calculus.

  • Differentiation finds the rate of change or slope of a function.

  • Integration finds the area under a curve or the accumulated value.

Both play a key role in physics, engineering, economics, and real-life applications. Let’s explore all important formulas with easy explanations.

Why it matters
Calculus uses derivatives in many contexts: motion (velocity), optimization (finding maxima/minima), curve-shape analysis, and more. Having the standard formulas at your fingertips makes solving problems much easier.


1. Basic Rules of Differentiation

These are the building blocks for all other derivative work:

  • Constant rule: If y=cy = cy=c (where ccc is a constant), then dydx=0. \frac{dy}{dx} = 0.dxdy​=0.

  • Power rule: If y=xny = x^ny=xn, then dydx=nxn1. \frac{dy}{dx} = n \cdot x^{\,n-1}.dxdy​=n⋅xn−1.

  • Constant multiple rule: If y=cf(x)y = c \cdot f(x)y=c⋅f(x), then y=cf(x). y' = c \cdot f'(x).y′=c⋅f′(x).

  • Sum/difference rule: If y=f(x)±g(x)y = f(x) \pm g(x)y=f(x)±g(x), then y=f(x)±g(x). y' = f'(x) \pm g'(x).y′=f′(x)±g′(x).

  • Product rule: If y=f(x)g(x)y = f(x)\cdot g(x)y=f(x)⋅g(x), then

    dydx=f(x)g(x)+f(x)g(x). \frac{dy}{dx} = f'(x)\cdot g(x) + f(x)\cdot g'(x).dxdy​=f′(x)⋅g(x)+f(x)⋅g′(x).
  • Quotient rule: If y=f(x)g(x)y = \frac{f(x)}{g(x)}y=g(x)f(x)​, then

    dydx=g(x)f(x)f(x)g(x)[g(x)]2. \frac{dy}{dx} = \frac{\,g(x)\,f'(x) - f(x)\,g'(x)\,}{[g(x)]^2}.dxdy​=[g(x)]2g(x)f′(x)−f(x)g′(x)​.

These rules let you handle combinations of functions. 


2. Derivatives of Common Functions

Once you know the rules, you also need the derivatives of commonly encountered functions:

Algebraic / Power / Log-exponential functions

  • ddx(ax)=axlna \frac{d}{dx}(a^x) = a^x \ln adxd​(ax)=axlna

  • ddx(ex)=ex \frac{d}{dx}(e^x) = e^xdxd​(ex)=ex

  • ddx(lnx)=1x \frac{d}{dx}(\ln x) = \frac1xdxd​(lnx)=x1​

  • ddx(logax)=1(lna)x\frac{d}{dx}(\log_a x) = \frac1{(\ln a)\,x}dxd​(loga​x)=(lna)x1​


Trigonometric functions

  • ddx(sinx)=cosx \frac{d}{dx}(\sin x) = \cos xdxd​(sinx)=cosx

  • ddx(cosx)=sinx \frac{d}{dx}(\cos x) = -\sin xdxd​(cosx)=−sinx

  • ddx(tanx)=sec2x \frac{d}{dx}(\tan x) = \sec^2 xdxd​(tanx)=sec2x

  • ddx(cotx)=csc2x \frac{d}{dx}(\cot x) = -\csc^2 xdxd​(cotx)=−csc2x

  • ddx(secx)=secxtanx \frac{d}{dx}(\sec x) = \sec x\,\tan xdxd​(secx)=secxtanx

  • ddx(cscx)=cscxcotx \frac{d}{dx}(\csc x) = -\csc x\,\cot xdxd​(cscx)=−cscxcotx


Inverse trigonometric functions

  • ddx(sin1x)=11x2\frac{d}{dx}(\sin^{-1} x) = \frac1{\sqrt{1 - x^2}}dxd​(sin−1x)=1−x2​1​

  • ddx(cos1x)=11x2\frac{d}{dx}(\cos^{-1} x) = -\frac1{\sqrt{1 - x^2}}dxd​(cos−1x)=−1−x2​1​

  • ddx(tan1x)=11+x2\frac{d}{dx}(\tan^{-1} x) = \frac1{1 + x^2}dxd​(tan−1x)=1+x21​

  • Other similar forms for sec1x\sec^{-1} xsec−1x, csc1x\csc^{-1} xcsc−1x, cot1x\cot^{-1} xcot−1x.


Hyperbolic functions (less common in early courses)

  • ddx(sinhx)=coshx \frac{d}{dx}(\sinh x) = \cosh xdxd​(sinhx)=coshx

  • ddx(coshx)=sinhx \frac{d}{dx}(\cosh x) = \sinh xdxd​(coshx)=sinhx

  • And similar derivative rules for tanhx\tanh xtanhx, cschx\mathrm{csch}\,xcschx, sechx\mathrm{sech}\,xsechx, cothx\coth xcothx.


3. Composite Functions: The Chain Rule

When you have a function inside another function—say y=f(g(x))y = f(g(x))y=f(g(x))—you use the chain rule:

dydx=f(g(x))g(x). \frac{dy}{dx} = f'(g(x)) \cdot g'(x).dxdy​=f′(g(x))⋅g′(x).

In more extended form: if you have several layers y=f(h(g(x)))y = f(h(g(x)))y=f(h(g(x))), then differentiate each layer from outer to inner:

dydx=f(h(g(x)))h(g(x))g(x). \frac{dy}{dx} = f'(h(g(x))) \cdot h'(g(x)) \cdot g'(x).dxdy​=f′(h(g(x)))⋅h′(g(x))⋅g′(x).

This rule is essential for many real-world problems and advanced functions.


4. Why Use the Formulas and How to Get Good at Them

  • Faster solving: With the standard formulas memorised, you can compute derivatives quickly instead of deriving each one from the limit definition.

  • Confidence: You’ll feel more comfortable with algebraic manipulations, especially rules like product/quotient, chain rule, implicit differentiation.

  • Deeper understanding: Knowing which rule to apply and why helps you understand what the derivative means (rise over run, slope of tangent, instantaneous rate) rather than blindly applying memorised steps.

  • Better in exams: Many exams ask for derivative of combined functions (e.g., ln(sinx) \ln(\sin x)ln(sinx), ex2cosxe^{x^2} \cdot \cos xex2⋅cosx). If you recognise the inner/outer structure and apply chain/product rules you’ll be more accurate.

Tip: Practice is key. Take a function, identify which rules apply (power + chain, or trig + product, etc), then compute step by step. Also verify your result by thinking: does the sign make sense? Is the domain valid (especially for inverse trig)?


Differentiation Formulas

No.FormulaDescription
1ddxk=0\frac{d}{dx} k = 0dxd​k=0Derivative of a constant
2ddx[f(x)±g(x)]=f(x)±g(x)\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)dxd​[f(x)±g(x)]=f′(x)±g′(x)Sum/Difference rule
3ddx[kf(x)]=kf(x)\frac{d}{dx}[k \cdot f(x)] = k \cdot f'(x)dxd​[k⋅f(x)]=k⋅f′(x)Constant multiple rule
4ddx[f(x)g(x)]=f(x)g(x)+g(x)f(x)\frac{d}{dx}[f(x) \cdot g(x)] = f(x)g'(x) + g(x)f'(x)dxd​[f(x)⋅g(x)]=f(x)g′(x)+g(x)f′(x)Product rule
5ddx[f(x)g(x)]=g(x)f(x)f(x)g(x)[g(x)]2\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}dxd​[g(x)f(x)​]=[g(x)]2g(x)f′(x)−f(x)g′(x)​Quotient rule
6ddxf(g(x))=f(g(x))g(x)\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)dxd​f(g(x))=f′(g(x))⋅g′(x)Chain rule
7ddxxn=nxn1\frac{d}{dx} x^n = n x^{n-1}dxd​xn=nxn−1Power rule
8ddxsinx=cosx\frac{d}{dx} \sin x = \cos xdxd​sinx=cosxTrigonometric
9ddxcosx=sinx\frac{d}{dx} \cos x = -\sin xdxd​cosx=−sinxTrigonometric
10ddxtanx=sec2x\frac{d}{dx} \tan x = \sec^2 xdxd​tanx=sec2xTrigonometric
11ddxcotx=csc2x\frac{d}{dx} \cot x = -\csc^2 xdxd​cotx=−csc2xTrigonometric
12ddxsecx=secxtanx\frac{d}{dx} \sec x = \sec x \tan xdxd​secx=secxtanxTrigonometric
13ddxcscx=cscxcotx\frac{d}{dx} \csc x = -\csc x \cot xdxd​cscx=−cscxcotxTrigonometric
14ddxex=ex\frac{d}{dx} e^x = e^xdxd​ex=exExponential
15ddxax=axlna\frac{d}{dx} a^x = a^x \ln adxd​ax=axlnaExponential
16( \frac{d}{dx} \lnx
17ddxsin1x=11x2\frac{d}{dx} \sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}dxd​sin−1x=1−x2​1​Inverse trig
18ddxcos1x=11x2\frac{d}{dx} \cos^{-1}x = -\frac{1}{\sqrt{1 - x^2}}dxd​cos−1x=−1−x2​1​Inverse trig
19ddxtan1x=11+x2\frac{d}{dx} \tan^{-1}x = \frac{1}{1 + x^2}dxd​tan−1x=1+x21​Inverse trig
20ddxcot1x=11+x2\frac{d}{dx} \cot^{-1}x = -\frac{1}{1 + x^2}dxd​cot−1x=−1+x21​Inverse trig
21( \frac{d}{dx} \sec^{-1}x = \frac{1}{x
22( \frac{d}{dx} \csc^{-1}x = -\frac{1}{x


Integration Formulas

No.FormulaDescription
1dx=x+C\int dx = x + C∫dx=x+CBasic integration
2xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C∫xndx=n+1xn+1​+CPower rule
3( \int \frac{dx}{x} = \lnx
4exdx=ex+C\int e^x dx = e^x + C∫exdx=ex+CExponential
5axdx=axlna+C\int a^x dx = \frac{a^x}{\ln a} + C∫axdx=lnaax​+CExponential
6lnxdx=xlnxx+C\int \ln x\, dx = x\ln x - x + C∫lnxdx=xlnx−x+CLogarithmic
7sinxdx=cosx+C\int \sin x\, dx = -\cos x + C∫sinxdx=−cosx+CTrigonometric
8cosxdx=sinx+C\int \cos x\, dx = \sin x + C∫cosxdx=sinx+CTrigonometric
9( \int \tan x, dx = -\ln\cos x
10( \int \cot x, dx = \ln\sin x
11( \int \sec x, dx = \ln\sec x + \tan x
12( \int \csc x, dx = -\ln\csc x + \cot x
13sec2xdx=tanx+C\int \sec^2 x\, dx = \tan x + C∫sec2xdx=tanx+CTrigonometric
14csc2xdx=cotx+C\int \csc^2 x\, dx = -\cot x + C∫csc2xdx=−cotx+CTrigonometric
15secxtanxdx=secx+C\int \sec x \tan x\, dx = \sec x + C∫secxtanxdx=secx+CTrigonometric
16cscxcotxdx=cscx+C\int \csc x \cot x\, dx = -\csc x + C∫cscxcotxdx=−cscx+CTrigonometric
17dxa2x2=sin1xa+C\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \frac{x}{a} + C∫a2−x2​dx​=sin−1ax​+CInverse trig
18dxa2+x2=1atan1xa+C\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C∫a2+x2dx​=a1​tan−1ax​+CInverse trig
19( \int \frac{dx}{x\sqrt{x^2 - a^2}} = \frac{1}{a} \sec^{-1} \frac{x


Final Thoughts

Keep a list of these formulas handy. Practice applying them. Start with simple cases (power, trig) and then move to harder ones (chain rule, quotient). With steady practice, you’ll get comfortable extracting derivatives quickly.

If you are looking to get into top IIT institutions, then cracking the JEE Main and Advanced should be your ultimate prior. With proper hands-on learning, practice, and curriculum, students at MNR Scottsdale are guided with the excellent preparation to meet these challenges and fulfill their dreams.

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