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%1.85\%
    Yearamp;2024amp;2023amp;2022amp;2021amp;2020amp;2019hline%amp;1.8amp;1.7amp;1.7amp;1.2amp;2.8amp;1.9hline\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%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

  • ddx(xn)=nxn−1    (n∈R)\frac{d}{dx}(x^n)=nx^{n-1}\;\;(n\in\mathbb R)

  • ddx(ex)=ex\frac{d}{dx}(e^{x})=e^x

  • ddx(ax)=axln⁡a\frac{d}{dx}(a^{x})=a^{x}\ln a

  • ddx(ln⁡∣x∣)=1x\frac{d}{dx}(\ln |x|)=\frac1x

  • ddx(log⁡a∣x∣)=1xln⁡a\frac{d}{dx}(\log_{a}|x|)=\frac1{x\ln a}

  • Trigonometry

    • ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x)=\cos x

    • ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x)= -\sin x

    • ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x)=\sec^{2}x

    • ddx(cot⁡x)=−csc⁡2x\frac{d}{dx}(\cot x)= -\csc^{2}x

    • ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x)=\sec x\tan x

    • ddx(csc⁡x)=−csc⁡xcot⁡x\frac{d}{dx}(\csc x)= -\csc x\cot x

Product, Quotient & Chain Rules

  • Product: ddx[u v]=u v′+v u′\frac{d}{dx}[u\,v]=u\,v'+v\,u'

  • Quotient: ddx(uv)=v u′−u v′v2\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{v\,u'-u\,v'}{v^{2}}

  • Chain: ddx[f(g(x))]=f′(g(x)) g′(x)\frac{d}{dx}[f(g(x))]=f'(g(x))\,g'(x)

Illustrative Problems

  1. y=extan⁡x  ⇒  y′=ex(tan⁡x+sec⁡2x)y=e^{x}\tan x\;\Rightarrow\;y' = e^{x}(\tan x + \sec^{2}x)

  2. y=ex1+sin⁡x  ⇒  y′=ex(1+sin⁡x−cos⁡x)(1+sin⁡x)2y=\dfrac{e^{x}}{1+\sin x}\;\Rightarrow\;y'=\dfrac{e^{x}(1+\sin x-\cos x)}{(1+\sin x)^{2}}

  3. Log form: y=xsin⁡xln⁡x  ⇒  y′=ln⁡x(xcos⁡x+sin⁡x)−xsin⁡xsin⁡x(ln⁡x)2y=\dfrac{x^{\sin x}}{\ln x}\;\Rightarrow\;y'=\dfrac{\ln x (x\cos x +\sin x)-x^{\sin x}\sin x}{(\ln x)^{2}}

  4. Three-function product question (JEE): given f(0)=1, g(0)=2, h(0)=?f(0)=1,\,g(0)=2,\,h(0)=? and pairwise derivatives, result (fgh)′(0)=32(fgh)'(0)=32 (worked by linearity of product rule).

Implicit Differentiation

General idea: treat yy as function of xx inside a relation F(x,y)=0F(x,y)=0.

Method 1 (direct): differentiate both sides, collect dydx\dfrac{dy}{dx}.
Method 2: solve for a variable when doable & substitute.

Examples

  1. x2+2xy+y3=4  ⇒  dydx=−2(x+y)2x+3y2x^{2}+2xy+y^{3}=4\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{-2(x+y)}{2x+3y^{2}}

  2. x+y=sin⁡(xy)  ⇒  dydx=ycos⁡(xy)−11−xcos⁡(xy)x+y=\sin(xy)\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{y\cos(xy)-1}{1-x\cos(xy)}

  3. Classic identity sin⁡y=xsin⁡(a+y)  ⇒  dydx=sin⁡asin⁡2(a+y)\sin y = x\sin(a+y)\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{\sin a}{\sin^{2}(a+y)}.

  4. Relation with roots x1+y+y1+x=0x\sqrt{1+y}+y\sqrt{1+x}=0 gives dydx=−1(1+x)2\dfrac{dy}{dx}=\dfrac{-1}{(1+x)^{2}} when y=x1+xy=\dfrac{x}{1+x}.

Parametric Differentiation

If x=g(t), y=f(t)x=g(t),\,y=f(t) then dydx=dy/dtdx/dt=f′(t)g′(t)\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{f'(t)}{g'(t)}.

  • x=at2, y=2at  ⇒  dydx=1tx=at^{2},\,y=2at\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{1}{t}.

  • x=acos⁡θ, y=bsin⁡θ  ⇒  dydx=−bacot⁡θx=a\cos\theta,\,y=b\sin\theta\;\Rightarrow\;\dfrac{dy}{dx}=-\dfrac{b}{a}\cot\theta.

  • Higher derivative example from JEE 2019: x=3tan⁡t, y=3sec⁡t  ⇒  d2ydx2∣t=π/4=12x=3\tan t,\,y=3\sec t\;\Rightarrow\;\left.\dfrac{d^{2}y}{dx^{2}}\right|_{t=\pi/4}=\dfrac{1}{2}.

Formula proof: d2ydx2=g′(t)f′′(t)−f′(t)g′′(t)[g′(t)]3\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

d ud v=du/dxdv/dx\frac{d\,u}{d\,v}=\frac{du/dx}{dv/dx}.

  • Example: derivative of sin⁡x\sin x w.r.t. tan⁡x\tan x equals cos⁡3x\cos^{3}x.

  • Given y=f(x3), z=g(x5), f′(x)=tan⁡x, g′(x)=sec⁡xy=f(x^{3}),\,z=g(x^{5}),\,f'(x)=\tan x,\,g'(x)=\sec x → dydz=3x2tan⁡(x3)5x4sec⁡(x5)=3tan⁡(x3)5x2sec⁡(x5)\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)

  1. y=xx⇒dydx=xx(1+ln⁡x)y=x^{x}\Rightarrow \dfrac{dy}{dx}=x^{x}(1+\ln x)

  2. y=xsin⁡x⇒dydx=xsin⁡x(sin⁡x ln⁡x+cos⁡x)y=x^{\sin x}\Rightarrow \dfrac{dy}{dx}=x^{\sin x}(\sin x\,\ln x+\cos x)

  3. Mixed power: 2x24x−1/(x2−1)3/22x^{2}\sqrt{4x-1}/(x^{2}-1)^{3/2} – after taking ln⁡\ln both sides, result
    y′=y(2xln⁡2+24x−1−3xx2−1)\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−1g=f^{-1}, then f(g(x))=x  ⟹  g′(x)=1f′(g(x))f(g(x))=x\implies g'(x)=\dfrac{1}{f'(g(x))}.

  • Example: f(x)=x3+ex2, g=f−1  ⇒  g′(1)=2f(x)=x^{3}+\dfrac{e^{x}}{2},\,g=f^{-1}\;\Rightarrow\;g'(1)=2.

  • JEE 2022: f(x)=x3+x−5, g=f−1⇒g′(63)=13(4)2+1=149f(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)=13g'(2)=\dfrac13 and h′(1)=666h'(1)=666.

Derivative of Inverse Trigonometric Functions (ITF)

  • ddx(sin⁡−1x)=11−x2\dfrac{d}{dx}(\sin^{-1}x)=\dfrac1{\sqrt{1-x^{2}}}

  • ddx(cos⁡−1x)=−11−x2\dfrac{d}{dx}(\cos^{-1}x)=-\dfrac1{\sqrt{1-x^{2}}}

  • ddx(tan⁡−1x)=11+x2\dfrac{d}{dx}(\tan^{-1}x)=\dfrac1{1+x^{2}}

  • ddx(cot⁡−1x)=−11+x2\dfrac{d}{dx}(\cot^{-1}x)= -\dfrac1{1+x^{2}}

  • ddx(sec⁡−1x)=1∣x∣x2−1\dfrac{d}{dx}(\sec^{-1}x)=\dfrac1{|x|\sqrt{x^{2}-1}}

  • ddx(csc⁡−1x)=−1∣x∣x2−1\dfrac{d}{dx}(\csc^{-1}x)= -\dfrac1{|x|\sqrt{x^{2}-1}}

Worked set:
(a) y=tan⁡−1!(1−cos⁡2x1+cos⁡2x)y=\tan^{-1}!\Bigl(\sqrt{\dfrac{1-\cos2x}{1+\cos2x}}\Bigr) requires quadrant splitting; derivative fails at x=0x=0.
(b) y=tan⁡−1!(cos⁡xsin⁡x)=−cot⁡−1(tan⁡x)y=\tan^{-1}!\bigl(\tfrac{\cos x}{\sin x}\bigr)=-\cot^{-1}(\tan x) gives y′=−12csc⁡2x2y'= -\dfrac12\csc^{2}\tfrac x2 etc.
(c) & (d) standard transformations to sin⁡−1(2x1−x2)\sin^{-1}(2x\sqrt{1-x^{2}}) etc. lead to compact y′=21−x2y'=\dfrac{2}{\sqrt{1-x^{2}}} or y′=−31−x2y'=- \dfrac{3}{\sqrt{1-x^{2}}} in given domains.

Infinite Radical / Series Forms

  • If y=sin⁡x+sin⁡x+…y=\sqrt{\sin x+\sqrt{\sin x+\ldots}}, then y2−y−sin⁡x=0y^{2}-y-\sin x=0 → y′=cos⁡x2y−1y'=\dfrac{\cos x}{2y-1}.

  • Classical y=x+y+x+…y=\sqrt{x+\sqrt{y+\sqrt{x+\ldots}}} eventually yields y2−y−2y−x=0y^{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⁡\ln to y2=xy2y^{2}=x^{y^{2}} giving y′=y3x [2−2y2ln⁡x]y' = \dfrac{y^{3}}{x\,[2-2y^{2}\ln x]}.

Higher-Order Derivatives & Special Identities

  • Given y=emx(ax+b)y=e^{mx}(ax+b) one proves
    d2ydx2−2mdydx+m2y=0\displaystyle\frac{d^{2}y}{dx^{2}}-2m\frac{dy}{dx}+m^{2}y=0.

  • IIT JEE 1998 identity: For cubic P(x)P(x) if y2=P(x)y^{2}=P(x) then P′′(x)P(x)=(P′(x))2P''(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)=202f'(10)=202.

Newton–Leibnitz Rule (Differentiation under the Integral Sign)

For F(x)=∫g(x)h(x)f(t) dtF(x)=\displaystyle\int_{g(x)}^{h(x)}f(t)\,dt:
F′(x)=f(h(x))h′(x)−f(g(x))g′(x)\boxed{F'(x)=f\bigl(h(x)\bigr)h'(x)-f\bigl(g(x)\bigr)g'(x)}.

Examples

  1. ddx∫0sin⁡xt3 dt=sin⁡3xcos⁡x−2x2\dfrac{d}{dx}\int_{0}^{\sin x} t^{3}\,dt = \sin^{3}x\cos x - 2x^{2}.

  2. ddx∫x22x!sin⁡t dt=2xsin⁡(x2)−2sin⁡(2x)\dfrac{d}{dx}\int_{x^{2}}^{2x}!\sin t\,dt = 2x\sin(x^{2})-2\sin(2x).

  3. ddx∫2x3ln⁡t dt=3x2ln⁡(x3)\dfrac{d}{dx}\int_{2}^{x^{3}}\ln t\,dt = 3x^{2}\ln(x^{3}).

  4. Advanced 2014 problem with matching derivatives gave F(2)=e4−1F(2)=e^{4}-1.

Miscellaneous Exam Highlights

  • Differentiation at points with absolute values: for y=x∣x∣y=x|x|, slope y′=2∣x∣y'=2|x| (non-existent at x=0x=0).

  • Algebraic-log mix (JEE Apr 2023): For 2x+3y=202^{x}+3^{y}=20, implicit derivative dydx=−2xln⁡23yln⁡3\dfrac{dy}{dx}= -\dfrac{2^{x}\ln2}{3^{y}\ln3} value at (2,2)(2,2) computed as −2+3ln⁡23+2ln⁡2-\dfrac{2+3\ln2}{3+2\ln2}.

  • Factorial-like product function f(x)=∏n=150(x−n)n(51−n)f(x)=\prod_{n=1}^{50}(x-n)^{n(51-n)} gave f′(51)f(51)=50×512=1275\dfrac{f'(51)}{f(51)}=\tfrac{50\times51}{2}=1275 so f′(51)5f(51)=255\frac{f'(51)}{5f(51)}=255.

  • Question “If x+y=3exx+y=3e^{x} find where ddx(xy)=0\dfrac{d}{dx}(x^{y})=0” ⇒ root at x=e2x=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=ax=a usually needs first solving f(b)=af(b)=a, then g′(a)=1/f′(b)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′=y′x′y'=\dfrac{y'}{x'} ; second order shown above.

  • Implicit: differentiate, group y′y' terms.

  • Inverse function: g′(a)=1f′(g(a))g'(a)=\dfrac1{f'(g(a))}.

  • Log differentiation: ln⁡y=ln⁡u⇒y′=y u′/u\ln y=\ln u \Rightarrow y' = y\,u'/u.

  • Newton-Leibnitz: [∫ghf]′=f(h)h′−f(g)g′\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!