Differentiability and the chain rule

What does differentiable mean?

In plain words: the slope of the graph must settle to one finite value whether you approach c from the left or from the right. A corner (two different one-sided slopes), a break, or a vertical tangent all stop a function from being differentiable there. The rules for sums, products and quotients, and the derivatives of x^ⁿ, sin x, cos x and tan x, are those of Class 11 (NCERT Table 5.3, page 119).

NCERT slip (page 119): at the end points of [a, b], NCERT calls the one-sided limits the “left hand derivative and right hand derivative of the function at a and b respectively”; at a it is the right-hand derivative and at b the left-hand one. In an exam, take each end point’s derivative from inside the interval: from the right at a, from the left at b.

Differentiable means continuous

If a function f is differentiable at a point c, then it is also continuous at that point.

NCERT Mathematics Part I (Class 12), Section 5.3, page 120

The proof (NCERT Theorem 3): for x ≠ c, f(x) − f(c) = [(f(x) − f(c))/(x − c)] × (x − c). As x → c, the bracket tends to f′(c) and the second factor to 0, so f(x) − f(c) → f′(c) × 0 = 0, that is, f(x) → f(c). So every differentiable function is continuous (NCERT Corollary 1).

The converse is false. f(x) = |x| is continuous at 0, but (|0 + h| − 0)/h = |h|/h is −1 for h < 0 and 1 for h > 0. The one-sided derivatives −1 and 1 differ, so |x| is not differentiable at 0 (NCERT page 120).

The chain rule

NCERT Theorem 4 (page 121): if f = v∘u, with t = u(x), and both dt/dx and dv/dt exist, then

NCERT’s illustration (pages 120–121): f(x) = (2x + 1)³ is t³ with t = 2x + 1, so df/dx = 3t² × 2 = 6(2x + 1)², the same as expanding to 8x³ + 12x² + 6x + 1 and differentiating. For three functions, f = (w∘u)∘v with t = v(x) and s = u(t), df/dx = (dw/ds)(ds/dt)(dt/dx) (NCERT page 121). NCERT Example 21: for f(x) = sin (x²), with t = x², df/dx = cos t × 2x = 2x cos (x²).

Implicit functions

When x and y are linked by an equation that is hard to solve for y, such as x + sin xy − y = 0, differentiate both sides with respect to x, treating y as a function of x: by the chain rule, d(y²)/dx = 2y dy/dx and d(sin y)/dx = cos y dy/dx. Then solve for dy/dx (NCERT Section 5.3.2, pages 122–123). A quotient obtained this way applies only where its denominator is non-zero. If the coefficient of dy/dx is zero, inspect the original curve and its branches instead of dividing by zero.

NCERT Example 23: for y + sin y = cos x, this gives dy/dx + cos y dy/dx = −sin x, so dy/dx = −sin x/(1 + cos y), for y ≠ (2n + 1)π.

Derivatives of inverse trigonometric functions

NCERT Example 24 (page 124): let y = sin^⁻¹ x, so x = sin y. Differentiating, 1 = cos y dy/dx. On the principal branch, y is between −π/2 and π/2, where cos y > 0, so cos y = √(1 − sin² y) = √(1 − x²). For −1 < x < 1:

At x = ±1, cos y = 0 and sin^⁻¹ x has no derivative. (The cos^⁻¹ result follows from cos^⁻¹ x = π/2 − sin^⁻¹ x.)

NCERT slip (page 124): in the table of these derivatives, the middle row is labelled “f^¹(x)” for f′(x), and the last row “Domain off” for “Domain of f′”.

Solved examples

Common mistakes to avoid

Quick revision summary

• f is differentiable at c when its left-hand and right-hand derivatives at c are finite and equal; a differentiable function is continuous, but |x| shows the converse fails.
• Chain rule: df/dx = (dv/dt)(dt/dx) for f = v∘u with t = u(x).
• Implicit functions: differentiate both sides in x, with d(g(y))/dx = g′(y) dy/dx, then solve for dy/dx.
• d/dx (sin^⁻¹ x) = 1/√(1 − x²), d/dx (cos^⁻¹ x) = −1/√(1 − x²) for −1 < x < 1, and d/dx (tan^⁻¹ x) = 1/(1 + x²).

Questions students ask

What is the difference between continuity and differentiability?

Continuity at a point means the graph has no break there (the limit of f(x) as x → c equals f(c)), while differentiability means it also has one finite slope there (the left-hand and right-hand derivatives are finite and equal), so differentiability is the stronger condition. Every function differentiable at c is continuous at c, but a continuous function can still fail to be differentiable, at a corner like that of |x| at 0 or at a vertical tangent: x^1/3 is continuous at 0, but its difference quotient h^1/3/h = 1/h^2/3 grows without bound as h → 0. A function with a break at a point, such as a jump, is never differentiable there.

What is the difference between the chain rule and the product rule?

The product rule differentiates a product of two functions, such as x² tan x, while the chain rule differentiates a function of a function, such as tan (x²): one adds two terms, (uv)′ = u′v + uv′, and the other multiplies the outer derivative, taken at the inside, by the inside’s derivative. So d/dx (x² tan x) = 2x tan x + x² sec² x, but d/dx tan (x²) = 2x sec² (x²). To choose, look at the last step in working out the function: a multiplication calls for the product rule, a function applied to an expression for the chain rule, and x tan (x²) needs both.

Why does implicit differentiation work, and when should you use it?

Implicit differentiation works because an equation such as x³ + y³ = 9, with y a differentiable function of x, holds for every x in an interval, so its two sides are equal functions of x and have equal derivatives; the chain rule turns the derivative of each term in y into its derivative with respect to y times dy/dx. Here 3x² + 3y² dy/dx = 0, so dy/dx = −x²/y², which is −1/4 at (1, 2), the same as differentiating y = (9 − x³)^1/3 directly. Use it when y is hard or impossible to isolate, as in x + sin xy − y = 0.

What are the derivatives of cot^⁻¹ x, sec^⁻¹ x and cosec^⁻¹ x?

The derivatives are d/dx (cot^⁻¹ x) = −1/(1 + x²) for every real x, d/dx (sec^⁻¹ x) = 1/(|x|√(x² − 1)) and d/dx (cosec^⁻¹ x) = −1/(|x|√(x² − 1)), both for |x| > 1. Each “co” function’s derivative is minus its partner’s, because cot^⁻¹ x = π/2 − tan^⁻¹ x and cosec^⁻¹ x = π/2 − sec^⁻¹ x on the principal branches. The |x| is there because sec^⁻¹ x rises on both parts of its domain, so its slope is positive for x < −1 too: at x = −2 it is 1/(2√3) ≈ 0.289. NCERT’s table (page 124) lists only sin^⁻¹, cos^⁻¹ and tan^⁻¹, while JEE Main Unit 7 names inverse trigonometric functions in general.

Are Rolle’s theorem and the mean value theorem in NCERT Class 12 and the JEE Main 2027 syllabus?

No, Rolle’s theorem and the mean value theorem are not in the 2026–27 NCERT book, whose Chapter 5 has only Theorems 1–5, nor in JEE Main 2027 Unit 7. The mean value theorem says: if f is continuous on [a, b] and differentiable on (a, b), then f′(c) = (f(b) − f(a))/(b − a) for some c in (a, b), such as c = √3 for x³ on [0, 3]; Rolle’s theorem is its special case f(a) = f(b), where f′(c) = 0. NCERT’s Chapter 6 still cites the mean value theorem as “Theorem 8 in Chapter 5” to prove its test for increasing and decreasing functions (page 154).

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Differentiate (i) sin (ax + b); (ii) sec (tan √x) (NCERT Exercise 5.2, Q3 and Q4, page 122).

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   (i) a cos (ax + b). (ii) The chain has three links: sec (tan √x) tan (tan √x) × sec² √x × 1/(2√x), for x > 0 wherever both secant factors are defined. At the domain endpoint x = 0, the quotient formula cannot be substituted directly; the right derivative is 1/2, since sec(tan √x) = 1 + x/2 + o(x).

2. Q2. Find dy/dx for (i) 2x + 3y = sin y; (ii) ax + by² = cos y (NCERT Exercise 5.3, Q2 and Q3, page 125).

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   (i) 2 + 3 dy/dx = cos y dy/dx, so dy/dx = 2/(cos y − 3). (ii) a + 2by dy/dx = −sin y dy/dx, so dy/dx = −a/(2by + sin y).

3. Q3. Find dy/dx if sin² x + cos² y = 1 (NCERT Exercise 5.3, Q8, page 125).

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   2 sin x cos x − 2 cos y sin y dy/dx = 0, so dy/dx = sin 2x/sin 2y where sin 2y ≠ 0. When both numerator and denominator vanish, this quotient gives no conclusion; the branches sin^² y = sin^² x must be examined separately.

4. Q4. Find dy/dx for y = cos^⁻¹ ((1 − x²)/(1 + x²)), 0 < x < 1 (NCERT Exercise 5.3, Q11, page 125).

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   With x = tan θ, 0 < θ < π/4, the bracket is cos 2θ with 0 < 2θ < π/2, so y = 2θ = 2 tan^⁻¹ x and dy/dx = 2/(1 + x²).