Differentiation – JEE Main & Advanced Comprehensive Bullet Notes
Nature of Chapter & Why It Matters
Differentiation is the first pillar of Calculus.
Creates foundation for:
Applications of Derivatives (AOD)
Indefinite & Definite Integration
Physics’ kinematics; however, JEE‐Math level is much deeper.
Format-based chapter → master each standard format by practice.
Instructor’s tip: “If you don’t get fluent here, AOD & Integration will hurt a lot.”
Exam Weightage (last 6 JEE cycles)
JEE-Main average weightage: 1.85\%
\begin{array}{|c|c|c|c|c|c|c|}\hline \text{Year}&2024&2023&2022&2021&2020&2019\\hline \text{\%}&1.8&1.7&1.7&1.2&2.8&1.9\\hline\end{array}JEE-Advanced average: 0.50\% (only 2023 had a 3 % hit).
Syllabus & Critical Sub-Topics
Standard derivatives list (algebraic, exponential, logarithmic, trigonometric).
Three Fundamental Rules
Product rule
Quotient rule
Chain rule ← marked “most important”.
Higher level skills
Implicit differentiation
Parametric differentiation
Differentiation of one function w.r.t. another
Logarithmic differentiation
Differentiation of inverse trigonometric functions (ITF)
Infinite series forms
Higher-order derivatives & Leibnitz theorem
Standard Derivatives You MUST Memorise
\frac{d}{dx}(x^n)=nx^{n-1}\;\;(n\in\mathbb R)
\frac{d}{dx}(e^{x})=e^x
\frac{d}{dx}(a^{x})=a^{x}\ln a
\frac{d}{dx}(\ln |x|)=\frac1x
\frac{d}{dx}(\log_{a}|x|)=\frac1{x\ln a}
Trigonometry
\frac{d}{dx}(\sin x)=\cos x
\frac{d}{dx}(\cos x)= -\sin x
\frac{d}{dx}(\tan x)=\sec^{2}x
\frac{d}{dx}(\cot x)= -\csc^{2}x
\frac{d}{dx}(\sec x)=\sec x\tan x
\frac{d}{dx}(\csc x)= -\csc x\cot x
Product, Quotient & Chain Rules
Product: \frac{d}{dx}[u\,v]=u\,v'+v\,u'
Quotient: \dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{v\,u'-u\,v'}{v^{2}}
Chain: \frac{d}{dx}[f(g(x))]=f'(g(x))\,g'(x)
Illustrative Problems
y=e^{x}\tan x\;\Rightarrow\;y' = e^{x}(\tan x + \sec^{2}x)
y=\dfrac{e^{x}}{1+\sin x}\;\Rightarrow\;y'=\dfrac{e^{x}(1+\sin x-\cos x)}{(1+\sin x)^{2}}
Log form: y=\dfrac{x^{\sin x}}{\ln x}\;\Rightarrow\;y'=\dfrac{\ln x (x\cos x +\sin x)-x^{\sin x}\sin x}{(\ln x)^{2}}
Three-function product question (JEE): given f(0)=1,\,g(0)=2,\,h(0)=? and pairwise derivatives, result (fgh)'(0)=32 (worked by linearity of product rule).
Implicit Differentiation
General idea: treat y as function of x inside a relation F(x,y)=0.
Method 1 (direct): differentiate both sides, collect \dfrac{dy}{dx}.
Method 2: solve for a variable when doable & substitute.
Examples
x^{2}+2xy+y^{3}=4\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{-2(x+y)}{2x+3y^{2}}
x+y=\sin(xy)\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{y\cos(xy)-1}{1-x\cos(xy)}
Classic identity \sin y = x\sin(a+y)\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{\sin a}{\sin^{2}(a+y)}.
Relation with roots x\sqrt{1+y}+y\sqrt{1+x}=0 gives \dfrac{dy}{dx}=\dfrac{-1}{(1+x)^{2}} when y=\dfrac{x}{1+x}.
Parametric Differentiation
If x=g(t),\,y=f(t) then \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{f'(t)}{g'(t)}.
x=at^{2},\,y=2at\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{1}{t}.
x=a\cos\theta,\,y=b\sin\theta\;\Rightarrow\;\dfrac{dy}{dx}=-\dfrac{b}{a}\cot\theta.
Higher derivative example from JEE 2019: x=3\tan t,\,y=3\sec t\;\Rightarrow\;\left.\dfrac{d^{2}y}{dx^{2}}\right|_{t=\pi/4}=\dfrac{1}{2}.
Formula proof: \displaystyle\frac{d^{2}y}{dx^{2}}=\frac{g'(t)f''(t)-f'(t)g''(t)}{\bigl[g'(t)\bigr]^{3}}.
Differentiation w.r.t. Another Function
\frac{d\,u}{d\,v}=\frac{du/dx}{dv/dx}.
Example: derivative of \sin x w.r.t. \tan x equals \cos^{3}x.
Given y=f(x^{3}),\,z=g(x^{5}),\,f'(x)=\tan x,\,g'(x)=\sec x → \dfrac{dy}{dz}=\dfrac{3x^{2}\tan(x^{3})}{5x^{4}\sec(x^{5})} = \dfrac{3\tan(x^{3})}{5x^{2}\sec(x^{5})}.
Logarithmic Differentiation (useful for power-to-power)
y=x^{x}\Rightarrow \dfrac{dy}{dx}=x^{x}(1+\ln x)
y=x^{\sin x}\Rightarrow \dfrac{dy}{dx}=x^{\sin x}(\sin x\,\ln x+\cos x)
Mixed power: 2x^{2}\sqrt{4x-1}/(x^{2}-1)^{3/2} – after taking \ln both sides, result
\displaystyle y'=y\left(2x\ln2+\frac{2}{4x-1}-\frac{3x}{x^{2}-1}\right) (simplified form shown in transcript).
Inverse Function Derivatives
Fundamental rule: If g=f^{-1}, then f(g(x))=x\implies g'(x)=\dfrac{1}{f'(g(x))}.
Example: f(x)=x^{3}+\dfrac{e^{x}}{2},\,g=f^{-1}\;\Rightarrow\;g'(1)=2.
JEE 2022: f(x)=x^{3}+x-5,\,g=f^{-1}\Rightarrow g'(63)=\dfrac{1}{3(4)^{2}+1}=\dfrac{1}{49}.
Advanced 2016 multi-function cascade produced g'(2)=\dfrac13 and h'(1)=666.
Derivative of Inverse Trigonometric Functions (ITF)
\dfrac{d}{dx}(\sin^{-1}x)=\dfrac1{\sqrt{1-x^{2}}}
\dfrac{d}{dx}(\cos^{-1}x)=-\dfrac1{\sqrt{1-x^{2}}}
\dfrac{d}{dx}(\tan^{-1}x)=\dfrac1{1+x^{2}}
\dfrac{d}{dx}(\cot^{-1}x)= -\dfrac1{1+x^{2}}
\dfrac{d}{dx}(\sec^{-1}x)=\dfrac1{|x|\sqrt{x^{2}-1}}
\dfrac{d}{dx}(\csc^{-1}x)= -\dfrac1{|x|\sqrt{x^{2}-1}}
Worked set:
(a) y=\tan^{-1}!\Bigl(\sqrt{\dfrac{1-\cos2x}{1+\cos2x}}\Bigr) requires quadrant splitting; derivative fails at x=0.
(b) y=\tan^{-1}!\bigl(\tfrac{\cos x}{\sin x}\bigr)=-\cot^{-1}(\tan x) gives y'= -\dfrac12\csc^{2}\tfrac x2 etc.
(c) & (d) standard transformations to \sin^{-1}(2x\sqrt{1-x^{2}}) etc. lead to compact y'=\dfrac{2}{\sqrt{1-x^{2}}} or y'=- \dfrac{3}{\sqrt{1-x^{2}}} in given domains.
Infinite Radical / Series Forms
If y=\sqrt{\sin x+\sqrt{\sin x+\ldots}}, then y^{2}-y-\sin x=0 → y'=\dfrac{\cos x}{2y-1}.
Classical y=\sqrt{x+\sqrt{y+\sqrt{x+\ldots}}} eventually yields y^{2}-y-\sqrt{2y}-x=0 then derivative by implicit manipulation (full algebra included in transcript).
Nested power y = (\sqrt[\sqrt[\sqrt{\;}]{}]{x})^{(\sqrt[\;]{x})^{\cdots}} boiled down via \ln to y^{2}=x^{y^{2}} giving y' = \dfrac{y^{3}}{x\,[2-2y^{2}\ln x]}.
Higher-Order Derivatives & Special Identities
Given y=e^{mx}(ax+b) one proves
\displaystyle\frac{d^{2}y}{dx^{2}}-2m\frac{dy}{dx}+m^{2}y=0.IIT JEE 1998 identity: For cubic P(x) if y^{2}=P(x) then P''(x)P(x)=\bigl(P'(x)\bigr)^{2} (demonstrated via differentiating 3 times).
JEE 2024 (27 Jan, S 1): Crafted polynomial with hidden constants leading to f'(10)=202.
Newton–Leibnitz Rule (Differentiation under the Integral Sign)
For F(x)=\displaystyle\int_{g(x)}^{h(x)}f(t)\,dt:
\boxed{F'(x)=f\bigl(h(x)\bigr)h'(x)-f\bigl(g(x)\bigr)g'(x)}.
Examples
\dfrac{d}{dx}\int_{0}^{\sin x} t^{3}\,dt = \sin^{3}x\cos x - 2x^{2}.
\dfrac{d}{dx}\int_{x^{2}}^{2x}!\sin t\,dt = 2x\sin(x^{2})-2\sin(2x).
\dfrac{d}{dx}\int_{2}^{x^{3}}\ln t\,dt = 3x^{2}\ln(x^{3}).
Advanced 2014 problem with matching derivatives gave F(2)=e^{4}-1.
Miscellaneous Exam Highlights
Differentiation at points with absolute values: for y=x|x|, slope y'=2|x| (non-existent at x=0).
Algebraic-log mix (JEE Apr 2023): For 2^{x}+3^{y}=20, implicit derivative \dfrac{dy}{dx}= -\dfrac{2^{x}\ln2}{3^{y}\ln3} value at (2,2) computed as -\dfrac{2+3\ln2}{3+2\ln2}.
Factorial-like product function f(x)=\prod_{n=1}^{50}(x-n)^{n(51-n)} gave \dfrac{f'(51)}{f(51)}=\tfrac{50\times51}{2}=1275 so \frac{f'(51)}{5f(51)}=255.
Question “If x+y=3e^{x} find where \dfrac{d}{dx}(x^{y})=0” ⇒ root at x=e^{2} via logarithmic differentiation.
Quick Ethical / Exam Tips
Always confirm domain when inverse or radical functions appear.
At non-smooth points (|x|, piecewise), check left & right derivative.
For inverse functions, computing derivative at x=a usually needs first solving f(b)=a, then g'(a)=1/f'(b).
In nested radicals/series, square once to build an algebraic relation; avoid infinite writing.
Newton–Leibnitz trick: think upper limit – lower limit, then multiply by their inside derivatives.
Ultra-Condensed Formula Sheet (one-glance)
Standard powers/exp/log/trig derivatives (see first table).
Rules: product, quotient, chain.
Parametric: y'=\dfrac{y'}{x'} ; second order shown above.
Implicit: differentiate, group y' terms.
Inverse function: g'(a)=\dfrac1{f'(g(a))}.
Log differentiation: \ln y=\ln u \Rightarrow y' = y\,u'/u.
Newton-Leibnitz: \bigl[\int_{g}^{h}f\bigr]' = f(h)h' - f(g)g'.
ITF derivatives list (6 items).
Happy practicing — master each “format”, then bombard past papers!