Class 12 — Mathematics (NCERT)

Continuity and Differentiability

Class 12

✓By the end of this lesson students will be able to define and examine the continuity of a function at a given point and on an interval.
✓By the end of this lesson students will be able to apply the chain rule effectively to differentiate composite functions.
✓By the end of this lesson students will be able to use logarithmic differentiation to find derivatives of functions involving products, quotients, and functions of the form [f(x)]^g(x).
✓By the end of this lesson students will be able to identify and apply appropriate differentiation techniques based on the structure of the given function.

Key concepts

Continuity of a Function at a Point

A function f(x) is said to be continuous at a point x = c if the function is defined at c, and the limit of the function as x approaches c exists and is equal to the value of the function at c. This means that the graph of the function has no break or jump at x = c.

A function f(x) is continuous at x = c if and only if: 1. f(c) is defined. 2. lim (x->c-) f(x) exists. 3. lim (x->c+) f(x) exists. 4. lim (x->c-) f(x) = lim (x->c+) f(x) = f(c).

Chain Rule

The chain rule is a formula to compute the derivative of a composite function. If y is a function of u, say y = f(u), and u is a function of x, say u = g(x), then y = f(g(x)) is a composite function. The chain rule states that the derivative of y with respect to x is the product of the derivative of y with respect to u and the derivative of u with respect to x.

If y = f(u) and u = g(x), then dy/dx = (dy/du) * (du/dx). Alternatively, d/dx [f(g(x))] = f'(g(x)) * g'(x).

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that are in the form [f(x)]^g(x) (a function raised to the power of another function), or functions that involve products or quotients of several functions. The process involves taking the natural logarithm of both sides of the equation, using logarithm properties to simplify, and then differentiating implicitly with respect to x.

If y = [f(x)]^g(x), then taking natural logarithm on both sides, log y = g(x) log f(x). Differentiating both sides with respect to x using implicit differentiation and product rule: (1/y) dy/dx = g'(x) log f(x) + g(x) * (1/f(x)) * f'(x). Hence, dy/dx = y [g'(x) log f(x) + g(x) * (f'(x)/f(x))].

Key facts to remember

1A function f(x) is continuous at x = c if lim (x->c-) f(x) = lim (x->c+) f(x) = f(c).
2All polynomial functions, trigonometric functions, exponential functions, and logarithmic functions are continuous in their respective domains.
3If f and g are continuous functions, then f+g, f-g, f*g, and f/g (where g(x) != 0) are also continuous.
4The Chain Rule: If y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).
5Logarithmic differentiation is used for functions of the form [f(x)]^g(x) or complex products/quotients.
6Remember standard derivatives: d/dx(x^n) = nx^(n-1), d/dx(sin x) = cos x, d/dx(cos x) = -sin x, d/dx(e^x) = e^x, d/dx(log x) = 1/x.

Worked examples

Example 1

Examine the continuity of the function f(x) defined by f(x) = { 2x+3, if x <= 2 ; 2x-3, if x > 2 } at x = 2.

IFirst, we find the value of the function at x = 2.

IIf(2) = 2(2) + 3 = 4 + 3 = 7.

IIINext, we find the Left Hand Limit (L.H.L.) at x = 2.

IVL.H.L. = lim (x->2-) f(x) = lim (h->0) f(2-h)

V= lim (h->0) [2(2-h) + 3] = lim (h->0) [4 - 2h + 3] = lim (h->0) [7 - 2h] = 7 - 2(0) = 7.

VINow, we find the Right Hand Limit (R.H.L.) at x = 2.

VIIR.H.L. = lim (x->2+) f(x) = lim (h->0) f(2+h)

VIII= lim (h->0) [2(2+h) - 3] = lim (h->0) [4 + 2h - 3] = lim (h->0) [1 + 2h] = 1 + 2(0) = 1.

9Since L.H.L. (7) is not equal to R.H.L. (1), the limit of f(x) as x approaches 2 does not exist.

10Therefore, the function f(x) is not continuous at x = 2.

Answer

The function f(x) is not continuous at x = 2.

For a function to be continuous at a point, L.H.L., R.H.L., and the function value at that point must all be equal.

Example 2

Differentiate sin(x^2 + 5) with respect to x.

ILet y = sin(x^2 + 5).

IILet u = x^2 + 5. Then y = sin(u).

IIIUsing the chain rule, dy/dx = (dy/du) * (du/dx).

IVFirst, find dy/du:

Vdy/du = d/du (sin u) = cos u.

VINext, find du/dx:

VIIdu/dx = d/dx (x^2 + 5) = 2x + 0 = 2x.

VIIISubstitute these back into the chain rule formula:

9dy/dx = (cos u) * (2x).

10Substitute u = x^2 + 5 back into the expression:

11dy/dx = 2x cos(x^2 + 5).

Answer

2x cos(x^2 + 5)

Always remember to differentiate the 'outer' function first, keeping the 'inner' function as is, and then multiply by the derivative of the 'inner' function.

Example 3

Differentiate x^x with respect to x.

ILet y = x^x.

IISince the variable is in the exponent, we take the natural logarithm on both sides.

IIIlog y = log (x^x).

IVUsing the logarithm property log(a^b) = b log a, we get:

Vlog y = x log x.

VINow, differentiate both sides with respect to x. We use implicit differentiation on the L.H.S. and the product rule on the R.H.S.

VIId/dx (log y) = d/dx (x log x).

VIII(1/y) dy/dx = (d/dx (x)) * (log x) + (x) * (d/dx (log x)).

9(1/y) dy/dx = (1) * (log x) + (x) * (1/x).

10(1/y) dy/dx = log x + 1.

11Multiply both sides by y to solve for dy/dx:

12dy/dx = y (log x + 1).

13Substitute y = x^x back into the expression:

14dy/dx = x^x (1 + log x).

Answer

x^x (1 + log x)

Logarithmic differentiation is particularly useful when the base and exponent are both functions of x.

Common mistakes

✗Failing to check all three conditions (L.H.L., R.H.L., and f(c)) or incorrectly evaluating limits for piecewise functions when examining continuity.
✗Forgetting to multiply by the derivative of the inner function when applying the chain rule, or incorrectly identifying the inner and outer functions.
✗Not applying implicit differentiation correctly on the L.H.S. (i.e., forgetting (1/y) dy/dx) during logarithmic differentiation, or making errors in applying logarithm properties.
✗Confusing the differentiation rules for d/dx(a^x) with d/dx(x^n) or d/dx(e^x).
✗Making algebraic errors in simplifying expressions after differentiation.

Exam tips

★For piecewise functions, always evaluate L.H.L., R.H.L., and f(c) separately and explicitly state your comparison to determine continuity.
★Break down complex composite functions into layers (e.g., y = f(u), u = g(v), v = h(x)) and apply the chain rule step-by-step: dy/dx = (dy/du) * (du/dv) * (dv/dx).
★Use logarithmic differentiation when you encounter functions of the form f(x)^g(x) or complicated products/quotients. Remember to take natural logarithm on both sides and then differentiate implicitly.
★Thoroughly practice differentiation of all standard functions and various combinations to build speed and accuracy for the examination.

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