Continuity at a point and on an interval, differentiability, the chain rule, implicit and parametric differentiation, derivatives of inverse trigonometric, exponential and logarithmic functions, logarithmic differentiation and second order derivatives. Our own step-by-step solutions to Exercises 5.1 to 5.7 and the Miscellaneous Exercise, one page per exercise, set out for step marks.
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What it tests. f is continuous at c when \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)\). Polynomials, sin, cos, \(e^x\), \(|x|\) are continuous everywhere; sums, products, quotients (where the denominator is non-zero) and composites of continuous functions are continuous. For a function defined in pieces, the only points to test are where the rule changes.
What it tests. Chain rule: \(\dfrac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)\), working from the outside in. A function is differentiable at c only if the left and right derivatives agree there; a corner (like \(|x - 1|\) at 1) or a jump (like \([x]\) at an integer) spoils this.
What it tests. Differentiate both sides with respect to x, treating y as a function of x (so \(\tfrac{d}{dx}y^2 = 2y\tfrac{dy}{dx}\)), then collect the \(\tfrac{dy}{dx}\) terms. For inverse trigonometric expressions, first simplify with a substitution (\(x = \tan\theta\), \(\sin\theta\), \(\cos\theta\)) inside the stated interval; standard results: \((\sin^{-1}x)' = \tfrac{1}{\sqrt{1 - x^2}}\), \((\cos^{-1}x)' = -\tfrac{1}{\sqrt{1 - x^2}}\), \((\tan^{-1}x)' = \tfrac{1}{1 + x^2}\).
What it tests. \((e^x)' = e^x\), \((\log x)' = \tfrac1x\) (x > 0), combined with the chain, product and quotient rules: \((e^{u})' = e^{u}u'\), \((\log u)' = \tfrac{u'}{u}\).
What it tests. For products, quotients and powers like \(u^v\) (variable base and exponent), take \(\log\) of both sides first: \(\log y = v\log u\), then \(\tfrac1y\tfrac{dy}{dx} = v'\log u + \tfrac{vu'}{u}\). For a sum such as \(x^x + (\sin x)^x\), differentiate each term separately (logs do not split sums).
What it tests. If \(x = f(t)\) and \(y = g(t)\), then \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\) wherever \(\tfrac{dx}{dt} \ne 0\); simplify the ratio with identities (half angles often help).
What it tests. Differentiate twice: \(y'' = \tfrac{d}{dx}(y')\). To prove a relation between \(y, y', y''\), it is often quicker to clear roots or fractions after the first derivative (e.g. \(\sqrt{1 - x^2}\,y' = 1\)) and differentiate that.
What it tests. Mixed practice: chain rule with powers, logarithmic differentiation for variable powers, simplifying inverse trigonometric expressions before differentiating, implicit and parametric second derivatives, and the link between continuity and differentiability.
Continuity and Differentiability has 40 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.
Also useful: free MCQs and case studies for Continuity and Differentiability · Class 12 formula sheet · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan
Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.