Continuity and Differentiability — Tests, Chain Rule, Implicit Differentiation

The Core Idea

Continuity and differentiability are about whether a function is “smooth” at a point. A continuous function has no breaks or jumps. A differentiable function has no sharp corners either — it’s smooth enough to have a tangent at every point.

This is one of the highest-weightage chapters in Class 12 maths. CBSE boards dedicate 8-10 marks to differentiation (this chapter + applications). JEE Main tests 1-2 questions on continuity/differentiability checks, and the chain rule is used in nearly every calculus problem.

Decision flowchart:

Is $f$ defined at $a$ ? If no — not continuous.
Does $\lim_{x \to a} f(x)$ exist? If no — not continuous.
Does the limit equal $f(a)$ ? If no — not continuous. If yes — continuous at $a$ .
Does LHD equal RHD? If no — not differentiable (corner or cusp). If yes — differentiable at $a$ (has a tangent line).

Key Terms & Definitions

Continuity at a point — A function $f$ is continuous at $x = a$ if: (1) $f(a)$ is defined, (2) $\lim_{x \to a} f(x)$ exists, and (3) $\lim_{x \to a} f(x) = f(a)$ .

Left-hand derivative (LHD) — $\lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$ .

Right-hand derivative (RHD) — $\lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$ .

Differentiability — $f$ is differentiable at $x = a$ if LHD = RHD. A differentiable function is always continuous, but a continuous function may not be differentiable (e.g., $|x|$ at $x = 0$ ).

Differentiation Rules

Function	Derivative
$x^n$	$nx^{n-1}$
$e^x$	$e^x$
$\ln x$	$\frac{1}{x}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$a^x$	$a^x \ln a$

If $y = f(g(x))$ , then:

\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

In words: differentiate the outer function (keeping inner as is), then multiply by the derivative of the inner function.

Product Rule: $(uv)' = u'v + uv'$

Quotient Rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$

When $y$ is not explicitly written as a function of $x$ , differentiate both sides w.r.t. $x$ , treating $y$ as a function of $x$ . Every time you differentiate a $y$ -term, multiply by $\frac{dy}{dx}$ .

For $y = f(x)^{g(x)}$ or products of many functions:

Take $\ln$ of both sides: $\ln y = g(x) \ln f(x)$
Differentiate both sides
Multiply both sides by $y$

Solved Examples — Easy to Hard

Example 1 (Easy — CBSE)

Differentiate $y = \sin(3x^2 + 5)$ .

By chain rule: $\frac{dy}{dx} = \cos(3x^2+5) \cdot 6x = \mathbf{6x\cos(3x^2+5)}$

Example 2 (Medium — CBSE)

Find $\frac{dy}{dx}$ if $x^2 + y^2 = 25$ .

Differentiating implicitly: $2x + 2y\frac{dy}{dx} = 0$

\frac{dy}{dx} = \mathbf{-\frac{x}{y}}

Example 3 (Medium — JEE Main)

Differentiate $y = x^{\sin x}$ .

Take $\ln$ : $\ln y = \sin x \cdot \ln x$ .

Differentiate: $\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}$

\frac{dy}{dx} = x^{\sin x}\left(\cos x \cdot \ln x + \frac{\sin x}{x}\right)

Example 4 (Hard — JEE Main)

Check continuity and differentiability of $f(x) = |x-1| + |x-2|$ at $x = 1$ and $x = 2$ .

For $x < 1$ : $f(x) = (1-x) + (2-x) = 3 - 2x$

For $1 \leq x < 2$ : $f(x) = (x-1) + (2-x) = 1$

For $x \geq 2$ : $f(x) = (x-1) + (x-2) = 2x - 3$

At $x = 1$ : LHL $= 3-2 = 1$ , RHL $= 1$ , $f(1) = 1$ . Continuous. LHD $= -2$ , RHD $= 0$ . Not differentiable (corner).

At $x = 2$ : LHL $= 1$ , RHL $= 1$ , $f(2) = 1$ . Continuous. LHD $= 0$ , RHD $= 2$ . Not differentiable (corner).

Exam-Specific Tips

CBSE Board: Expect 4-6 marks from this chapter. Typical questions: check continuity at a point for a piecewise function, differentiate using chain rule, implicit differentiation, or logarithmic differentiation. Rolle’s theorem and LMVT also appear.

JEE Main: Continuity + differentiability of piecewise and modulus functions is tested regularly. Also: finding $f'(x)$ for composite or parametric functions, and questions where differentiability implies some condition on parameters.

Common Mistakes to Avoid

Mistake 1 — Assuming continuous implies differentiable. $|x|$ is continuous at $x = 0$ but not differentiable there. Differentiability is a stronger condition.

Mistake 2 — Forgetting the chain rule. When differentiating $\sin(5x)$ , students write $\cos(5x)$ instead of $5\cos(5x)$ . Always multiply by the derivative of the inner function.

Mistake 3 — Missing $dy/dx$ in implicit differentiation. Every time you differentiate a term involving $y$ , you must attach $dy/dx$ . Missing it makes the entire solution wrong.

Mistake 4 — Wrong sign in the quotient rule. It’s $u'v - uv'$ (numerator derivative first, then subtract). Swapping gives the wrong sign.

Mistake 5 — Not checking both LHD and RHD at suspect points. For piecewise functions, you must check left and right limits AND left and right derivatives at every breakpoint.

Practice Questions

Q1. Differentiate $y = e^{\tan x}$ .

$dy/dx = e^{\tan x} \cdot \sec^2 x$ .

Q2. Is $f(x) = \frac{|x|}{x}$ continuous at $x = 0$ ?

$f(x) = 1$ for $x > 0$ and $f(x) = -1$ for $x < 0$ . $f(0)$ is undefined. Not continuous (function not defined at 0, and limits from left and right don’t match).

Q3. Find $dy/dx$ if $\sin^2 x + \cos^2 y = 1$ .

$2\sin x \cos x + 2\cos y(-\sin y)\frac{dy}{dx} = 0$ . So $\frac{dy}{dx} = \frac{\sin 2x}{2\sin y \cos y} = \frac{\sin 2x}{\sin 2y}$ .

Q4. Differentiate $y = (\ln x)^x$ .

$\ln y = x \ln(\ln x)$ . Differentiate: $\frac{1}{y}\frac{dy}{dx} = \ln(\ln x) + \frac{x}{x \ln x} = \ln(\ln x) + \frac{1}{\ln x}$ . So $\frac{dy}{dx} = (\ln x)^x \left[\ln(\ln x) + \frac{1}{\ln x}\right]$ .

Q5. Find the value of $k$ if $f(x) = kx + 1$ for $x \leq 5$ and $f(x) = 3x - 5$ for $x > 5$ , given that $f$ is continuous at $x = 5$ .

Continuity requires $k(5) + 1 = 3(5) - 5 = 10$ . So $5k = 9$ , giving $k = 9/5$ .

Q6. Differentiate $y = \sin^{-1}(2x\sqrt{1-x^2})$ .

Put $x = \sin\theta$ . Then $2\sin\theta\cos\theta = \sin 2\theta$ . So $y = 2\sin^{-1}x$ . $dy/dx = \frac{2}{\sqrt{1-x^2}}$ (for $|x| \leq 1/\sqrt{2}$ ).

Q7. If $x = a\cos^3\theta$ , $y = a\sin^3\theta$ , find $dy/dx$ .

$dx/d\theta = -3a\cos^2\theta \sin\theta$ . $dy/d\theta = 3a\sin^2\theta \cos\theta$ . $dy/dx = \frac{3a\sin^2\theta \cos\theta}{-3a\cos^2\theta \sin\theta} = -\tan\theta$ .

Q8. Verify Rolle’s theorem for $f(x) = x^2 - 5x + 6$ on $[2, 3]$ .

$f(2) = 4 - 10 + 6 = 0$ , $f(3) = 9 - 15 + 6 = 0$ . $f$ is polynomial so continuous and differentiable. $f'(x) = 2x - 5 = 0$ gives $x = 5/2 \in (2, 3)$ . Rolle’s theorem verified.

FAQs

Can a function be differentiable but not continuous?

No. Differentiability at a point requires continuity at that point. If a function is differentiable, it’s automatically continuous.

What does the derivative represent geometrically?

The derivative $f'(a)$ is the slope of the tangent line to the curve $y = f(x)$ at the point $x = a$ .

When do we use logarithmic differentiation?

When the function has a variable in both the base and the exponent ( $x^x$ , $x^{\sin x}$ ), or when differentiating a product of many functions.

What is the second derivative test?

$f''(a) > 0$ means the function has a local minimum at $a$ . $f''(a) < 0$ means a local maximum. $f''(a) = 0$ is inconclusive.

What is Rolle’s theorem?

If $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $f(a) = f(b)$ , then there exists at least one $c \in (a, b)$ where $f'(c) = 0$ .

Additional Concepts

Mean Value Theorem (LMVT)

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , there exists $c \in (a,b)$ such that:

f'(c) = \frac{f(b) - f(a)}{b - a}

Geometrically: the tangent at some point $c$ is parallel to the secant joining $(a, f(a))$ and $(b, f(b))$ .

Rolle’s theorem is LMVT with $f(a) = f(b)$ — the secant is horizontal, so the tangent is too.

Higher-order derivatives

The second derivative $f''(x)$ tells us about concavity:

$f''(x) > 0$ : concave up (curve bends upward, like a cup)
$f''(x) < 0$ : concave down (curve bends downward, like a cap)
$f''(x) = 0$ : possible inflection point (concavity changes)

Derivatives of inverse trig functions

Function	Derivative
$\sin^{-1}x$	$\frac{1}{\sqrt{1-x^2}}$
$\cos^{-1}x$	$\frac{-1}{\sqrt{1-x^2}}$
$\tan^{-1}x$	$\frac{1}{1+x^2}$
$\cot^{-1}x$	$\frac{-1}{1+x^2}$
$\sec^{-1}x$	$\frac{1}{
$\csc^{-1}x$	$\frac{-1}{

Parametric differentiation

If $x = f(t)$ and $y = g(t)$ :

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}

Example: $x = a\cos\theta$ , $y = a\sin\theta$ .

$\frac{dy}{dx} = \frac{a\cos\theta}{-a\sin\theta} = -\cot\theta$

Additional Practice Questions

Q9. Verify LMVT for $f(x) = x^2 + 2x + 3$ on $[1, 4]$ .

$f(1) = 6$ , $f(4) = 27$ . $\frac{f(4)-f(1)}{4-1} = \frac{21}{3} = 7$ . $f'(x) = 2x + 2 = 7 \implies x = 5/2 = 2.5 \in (1,4)$ . Verified.

Q10. Differentiate $y = \tan^{-1}\frac{2x}{1-x^2}$ .

Recognise: $\tan^{-1}\frac{2x}{1-x^2} = 2\tan^{-1}x$ (for $|x| < 1$ ). So $\frac{dy}{dx} = \frac{2}{1+x^2}$ .

Summary of When to Use Each Differentiation Technique

Technique	When to use	Example
Chain rule	Composite functions	$\sin(e^x)$ , $\ln(\cos x)$
Product rule	Product of two functions	$x^2 \sin x$ , $e^x \ln x$
Quotient rule	Ratio of two functions	$\frac{\sin x}{x^2 + 1}$
Implicit	$y$ not isolated	$x^2 + y^2 = 1$
Logarithmic	Variable base and exponent	$x^x$ , $(\sin x)^{\cos x}$
Parametric	$x = f(t)$ , $y = g(t)$	$x = a\cos t$ , $y = b\sin t$

When you see a function and cannot decide which rule to use, ask: is it a composition (chain), a product (product rule), a ratio (quotient rule), or does it have the variable in both base and exponent (logarithmic)? This decision tree covers 95% of differentiation problems.