2026?27 Syllabus Focus
The course structure groups marks by unit, but the paper design does not guarantee a fixed allocation for any chapter. These checkpoints follow the current syllabus; they are not a mark prediction.
- Continuity, differentiability and the chain rule for composite functions.
- Derivatives of inverse trigonometric, implicit, exponential and logarithmic functions; logarithmic differentiation.
- Parametric differentiation and second-order derivatives.
Last 24-Hour Checklist
Continuity (The Unbroken Path)
Imagine drawing a graph without lifting your pen. That's Continuity!
If there is a break, hole, or jump, it is Discontinuous.
The "Chalk Logic" (Limits)
For a function to be continuous at a point $x=c$:
Left Hand Limit = Right Hand Limit = Value of Function
$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$
Polynomials, Sine, Cosine, Exponential, and Constant functions are continuous everywhere in their domain.
Answer: No, because $f(0)$ is not defined.
Differentiability (Smooth Turns)
Differentiable means "No Sharp Corners".
A ball rolling on the graph should turn smoothly, not bounce off a sharp edge.
The Condition
LHD = RHD
Left Hand Derivative = Right Hand Derivative (at a point)
The Relation
Differentiable $\implies$ Continuous
But Continuous $\nRightarrow$ Differentiable.
Example: $|x|$ at $x=0$ is continuous (no break) but not differentiable (sharp V-shape).
Answer: No, it has a sharp corner (V-shape) at $x=1$.
High-Yield Parametric & Logarithmic Differentiation Rules
- Parametric Second Order Derivative Formula (The #1 Board Trap):
If $x = f(t)$ and $y = g(t)$, then $\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$.
The second derivative is: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$$ Caution: $\frac{d^2y}{dx^2} \neq \frac{d^2y/dt^2}{d^2x/dt^2}$. Forgetting the factor $\frac{dt}{dx} = \frac{1}{dx/dt}$ is the most common error in Board exams! - Logarithmic Differentiation for Sums ($u + v$) Trap:
When differentiating functions of the form $y = x^{\sin x} + (\sin x)^x$:
• You CANNOT take $\log$ on both sides directly because $\log(u + v) \neq \log u + \log v$!
• Correct method: Let $u = x^{\sin x}$ and $v = (\sin x)^x$, so $y = u + v \implies \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$. Differentiate $u$ and $v$ individually using logarithms. - Differentiability of Modulus Function $|x - a|$:
• $f(x) = |x - a|$ is continuous everywhere on $\mathbb{R}$.
• It is NOT differentiable at the corner point $x = a$ because $\text{LHD} = -1 \neq \text{RHD} = +1$.
• For $x \neq a$: $\frac{d}{dx}|x - a| = \frac{x - a}{|x - a|} = \frac{|x - a|}{x - a}$. - Differentiability Implies Continuity (Fundamental Theorem):
If a function is differentiable at $x = c$, it is ALWAYS continuous at $x = c$. However, the converse is FALSE (e.g. $f(x) = |x|$ at $x=0$). - 📌 CBSE Rationalized Syllabus Notice: Rolle's Theorem and Lagrange's Mean Value Theorem (NCERT Ex 5.8) have been deleted from the active CBSE Class 12 Board syllabus.
Standard Derivatives Table
| Function $f(x)$ | Derivative $f'(x)$ | Function $f(x)$ | Derivative $f'(x)$ |
|---|---|---|---|
| $x^n$ | $nx^{n-1}$ | $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ | $\tan x$ | $\sec^2 x$ |
| $\cot x$ | $-\csc^2 x$ | $\sec x$ | $\sec x \tan x$ |
| $\csc x$ | $-\csc x \cot x$ | $\sin^{-1} x$ | $\frac{1}{\sqrt{1-x^2}}$ |
| $\cos^{-1} x$ | $\frac{-1}{\sqrt{1-x^2}}$ | $\tan^{-1} x$ | $\frac{1}{1+x^2}$ |
| $\cot^{-1} x$ | $\frac{-1}{1+x^2}$ | $\sec^{-1} x$ | $\frac{1}{|x|\sqrt{x^2-1}}$ |
| $\csc^{-1} x$ | $\frac{-1}{|x|\sqrt{x^2-1}}$ | $a^x$ | $a^x \log_e a$ |
| $e^x$ | $e^x$ | $\log_e x$ | $1/x$ $(x > 0)$ |
| $\log_a x$ | $\frac{1}{x \log_e a}$ | $|x|$ | $x/|x|$, $x \neq 0$ |
Some Properties & Rules of Derivatives
Sum/Difference Rule
$(u \pm v)' = u' \pm v'$
Product Rule
$(uv)' = u'v + uv'$
Quotient Rule
$(\frac{u}{v})' = \frac{u'v - uv'}{v^2}, v \neq 0$
Chain Rule (Onion Peeling)
$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$
Differentiate outer layer first $\to$ inner layer.
Quick Check: Derivative of
$\sin(x^3)$?
$\cos(x^3) \cdot 3x^2$.
Use when function is of form $y = [u(x)]^{v(x)}$ (Variable Power).
Trick: Take $\log$ to bring the power down.
$\log y = v(x) \log[u(x)]$
Quick Check: Derivative of $x^x$?
$x^x(1 + \log x)$.
Special Differentiation Types
1. Implicit Function
Functions where $x$ and $y$ are mixed and $y$ cannot be explicitly expressed in terms of $x$. Differentiate both sides w.r.t $x$ and solve for $dy/dx$.
2. Parametric Function
When both variables $x$ and $y$ are expressed in terms of a third variable, say $t$ (the parameter), if $x = f(t)$ and $y = g(t)$, then:
Quick Check: $x=at^2, y=2at$.
$dy/dt = 2a$, $dx/dt = 2at$.
$dy/dx = (2a)/(2at) = 1/t$.
3. Second Order Derivative
Let $y = f(x)$, then $\frac{dy}{dx} = f'(x)$.
Differentiating again: $\frac{d}{dx}(\frac{dy}{dx}) = \frac{d^2y}{dx^2} = f''(x)$
Solved Examples (Board Marking Scheme)
Q1. Find $dy/dx$ if $y = \cos(\sin x^2)$. (2 Marks)
$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \frac{d}{dx}(\sin x^2)$.
$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \cos x^2 \cdot 2x$.
$\frac{dy}{dx} = -2x \cos x^2 \sin(\sin x^2)$.
Q2. If $y = e^x \sin x$, prove that $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$. (4 Marks)
$\frac{dy}{dx} = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$.
$\frac{d^2y}{dx^2} = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.
LHS: $2e^x \cos x - 2[e^x(\sin x + \cos x)] + 2(e^x \sin x)$
$= 2e^x \cos x - 2e^x \sin x - 2e^x \cos x + 2e^x \sin x = 0$. (Hence Proved)
Previous Year Questions (PYQs)
Ans: $x^2+x-6 = (x+3)(x-2) = 0 \Rightarrow x = -3, 2$.
Ans: Let $u = \sin^2 x, v = e^{\cos x}$. $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2\sin x \cos x}{-e\cos x \sin x} = \frac{-2\cos x}{e\cos x}$.
Hint: $y_1 = 2\tan^{-1} x \cdot \frac{1}{x^2+1}$. $(x^2+1)y_1 = 2\tan^{-1} x$. Differentiate again.
Exam Strategy & Mistake Bank
Common Mistakes
Scoring Tips
Practice Problems (Self-Assessment)
Level 1: Basic (1 Mark Each)
Q1. Find $dy/dx$ if $y = \log_{10} x$.
Direct differentiation is for $\log_e x$. So convert base first.
$y = \frac{\log_e x}{\log_e 10}$. (Constant $\frac{1}{\log_e 10}$ comes out)
$\frac{dy}{dx} = \frac{1}{\log_e 10} \cdot \frac{d}{dx}(\log_e x)$
$\frac{dy}{dx} = \frac{1}{\log_e 10} \cdot \frac{1}{x} = \frac{1}{x \log_e 10}$.
Q2. Is $f(x) = |x|$ continuous at $x = 0$?
LHL ($\lim_{x \to 0^-} -x$) = 0.
RHL ($\lim_{x \to 0^+} x$) = 0.
$f(0) = |0| = 0$.
Since LHL = RHL = $f(0) = 0$, the function is Continuous.
Level 2: Intermediate (2-3 Marks Each)
Q3. Differentiate $(\sin x)^x$ with respect to $x$.
Let $y = (\sin x)^x$. Take log on both sides.
$\log y = x \log(\sin x)$.
$\frac{1}{y} \frac{dy}{dx} = x · \frac{d}{dx}(\log \sin x) + \log \sin x · \frac{d}{dx}(x)$
$\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{\sin x} \cdot \cos x + \log \sin x \cdot 1$
$\frac{1}{y} \frac{dy}{dx} = x \cot x + \log \sin x$.
Multiply by $y$ on RHS.
$\frac{dy}{dx} = (\sin x)^x [x \cot x + \log \sin x]$.
Q4. Find $dy/dx$ if $x = a\cos θ, y = b\sin θ$.
$\frac{dx}{dθ} = a(-\sin θ) = -a\sin θ$.
$\frac{dy}{dθ} = b(\cos θ) = b\cos θ$.
$\frac{dy}{dx} = \frac{dy/dθ}{dx/dθ} = \frac{b\cos θ}{-a\sin θ}$.
$\frac{dy}{dx} = -\frac{b}{a} \cot θ$.
Level 3: Advanced (5 Marks Each)
Q5. If $y = \sin^{-1} x$, show that $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0$.
$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$.
Cross multiply: $\sqrt{1-x^2} y_1 = 1$.
Square both sides to remove root (makes differentiation easier).
$(1-x^2) y_1^2 = 1$.
Use Product Rule on LHS.
$(1-x^2) \cdot \frac{d}{dx}(y_1^2) + y_1^2 \cdot \frac{d}{dx}(1-x^2) = 0$
$(1-x^2) \cdot (2y_1 y_2) + y_1^2 \cdot (-2x) = 0$.
Divide entire equation by $2y_1$.
$(1-x^2)y_2 - x y_1 = 0$.
Hence Proved.
Chapter 5 Board Mastery Test
Practice test • 10 marks • 25 min