Continuity, Differentiability and Differentiation | ISC Class 12 Maths Notes

This note covers continuity at a point and on an interval, removable discontinuities, differentiability, rules of differentiation, composite and implicit functions, inverse trigonometric derivatives, exponential and logarithmic functions, logarithmic differentiation, parametric derivatives and second order derivatives.

What does continuity at a point mean?

A function assigns one output to each permitted input. Write its rule as y = f(x), where x is the input, f names the function and y is its output. Its domain is the set of permitted inputs; its range is the set of outputs.

Let c denote a fixed real number in the domain. The limit of f(x) as x approaches c describes the value approached by the outputs near c. The notation lim as x → c means “limit as x approaches c”. It does not mean simply substituting c.

Definition: A function f is continuous at c when its limit as x approaches c equals f(c), its actual value at c: $\lim_{x\to c}f(x)=f(c)$.

How do the two sides enter the test?

The left hand limit uses inputs smaller than c; the right hand limit uses inputs larger than c. Superscript − and + in c⁻ and c⁺ indicate these directions. At an interior point, both limits must exist and equal f(c).

1. Find the actual value f(c) using the part of the definition that includes c.
2. Find the left hand limit using the rule valid immediately to the left.
3. Find the right hand limit using the rule valid immediately to the right.
4. Compare all three quantities. Equal finite values establish continuity at c.

Worked example 1. Check whether f(x) = 2x + 3 is continuous at x = 1.

Answer: f(1) = 2(1) + 3 = 5. Also, the limit of 2x + 3 as x approaches 1 is 5. The limit equals the function value, so f is continuous at 1.

A function is discontinuous at a point of its domain when the continuity condition fails there. An existing function value alone is insufficient: the neighbouring values must approach that same value.

How are continuity on intervals and algebraic combinations tested?

An interval contains every real number between its endpoints. Let a and b be endpoints with a < b. The open interval (a, b) excludes both endpoints; the closed interval [a, b] includes both. Continuity on an interval requires continuity throughout it.

For a function defined on [a, b], use the right hand limit at a and the left hand limit at b. Inputs beyond these endpoints are outside the stated domain. At every interior point, use the ordinary two-sided condition.

Theorem: Algebra of continuous functions

Let f and g be real functions continuous at c. Their sum f + g, difference f − g and product fg are continuous at c. Their quotient f/g is continuous at c provided g(c) ≠ 0; ≠ means “not equal to”.

For the sum, the limit of f(x) + g(x) equals the sum of the separate limits. Continuity replaces these limits by f(c) and g(c). Their sum is exactly the value of f + g at c, which proves the required equality.

A polynomial is a finite sum of constant multiples of non-negative integer powers of x. Every polynomial is continuous at every real number. A rational function is a quotient of polynomials and is continuous wherever its denominator is non-zero.

Theorem: Continuity of a composite function

A composite function applies one function after another: (f ∘ g)(x) means f(g(x)). If g is continuous at c and f is continuous at g(c), then f ∘ g is continuous at c, provided the composition is defined.

Worked example 2. Discuss continuity of f(x) = 1/x, with domain x ≠ 0.

Answer: At any non-zero c, the limit of 1/x is 1/c = f(c). Thus f is continuous at every point of its domain. No real value assigned at zero can make it continuous there.

The phrase continuous function means continuous throughout its domain. It does not require the domain to contain every real number. Keep this distinction when a denominator excludes an input.

What is a removable discontinuity?

A removable discontinuity occurs when a finite limit exists at an input but the function value is missing or disagrees with that limit. The defect can be removed by defining or redefining the value to equal the limit.

These are two useful cases: a missing value, often called a hole, and an incorrectly assigned value. For a missing value, one is extending the original function to include the omitted input. The original function can still be continuous on its original domain.

Worked example 3. Let f(x) = x³ + 3 for x ≠ 0 and f(0) = 1. Test continuity and identify the value needed at zero.

Answer: The limit as x approaches 0 is 3, but f(0) = 1. Hence f is discontinuous at zero. Redefining f(0) as 3 makes the function continuous there without changing any other value.

How is a missing value repaired?

Consider f(x) = x + 2 for x < 0 and f(x) = −x + 2 for x > 0. No value is given at zero. Both one-sided limits equal 2, so extending the definition by f(0) = 2 joins the two parts continuously.

By contrast, a jump discontinuity has different finite one-sided limits. Changing one point cannot make these limits agree. For f(x) = 1 when x ≤ 0 and f(x) = 2 when x > 0, the limits at zero are 1 and 2.

The horizontal line y = 1 has an open circle at its intersection with the vertical axis. A separate filled point is labelled (0, 2), showing the assigned value at zero.

See Fig. 5.2 in your NCERT textbook

An open circle indicates an excluded graph point; a filled point indicates an included one. The horizontal line approaches height 1 from both sides, while the filled point gives value 2. Replacing that assigned value by 1 removes this discontinuity.

How does differentiability differ from continuity?

The derivative measures the limiting ratio of a change in output to a change in input. Let h be a non-zero input increment. The expression [f(c + h) − f(c)]/h is the difference quotient at c.

Definition: The derivative is $f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}$, provided this limit exists as a finite real number. The prime in f′ denotes differentiation. A function with such a derivative is differentiable at c.

The left hand derivative takes h approaching zero through negative values. The right hand derivative takes positive values. Differentiability at an interior point requires these two limits to be finite and equal. Differentiation is the process of finding a derivative.

A function is differentiable on an open interval when it is differentiable at every point of that interval. For a closed interval, endpoint derivatives, when required, use the right hand limit at the left endpoint and the left hand limit at the right endpoint.

Theorem: Differentiability implies continuity

If f is differentiable at c, it is continuous there. For h ≠ 0, write f(c + h) − f(c) as the difference quotient multiplied by h. As h approaches zero, this product approaches f′(c) × 0 = 0, proving continuity.

The converse is false. The modulus |x|, or absolute value, equals −x for x < 0 and x for x ≥ 0. Both sides approach zero, so |x| is continuous at zero, but its one-sided derivatives differ.

Worked example 4. Test differentiability of f(x) = |x| at zero.

Answer: Since f(0) = 0, the difference quotient is |h|/h. For h < 0 it equals −1; for h > 0 it equals 1. These limits differ, so f′(0) does not exist.

What happens for the greatest integer function?

The greatest integer function [x] is the largest integer not exceeding x. At an integer n, its left hand limit is n − 1, while its right hand limit and value are n. It is discontinuous, and therefore not differentiable, at every integer.

Between consecutive integers, [x] is constant. It is continuous and has derivative zero at every non-integer input. This is a local statement: each such input has a small surrounding interval on which the value remains unchanged.

Horizontal steps have filled left endpoints and open right endpoints. The steps occur at integer heights, with a jump between successive steps.

See Fig. 5.8 in your NCERT textbook

Which basic differentiation rules should be used?

Write dy/dx or y′ for the derivative of y with respect to x; d/dx denotes the operation of differentiating. In the following rules, u and v are differentiable functions of x, and u′ and v′ denote their derivatives.

How do sums, products and quotients differ?

A constant does not vary with x, so its derivative is zero. The power rule gives d(xⁿ)/dx = nxⁿ⁻¹ for a positive integer n. Differentiate a polynomial term by term using this rule and the constant multiple rule.

Trigonometric functions relate an angle to a unit circle, a circle of radius 1 centred at the origin: cos x and sin x are the horizontal and vertical coordinates of the point reached by the angle. Use radians, the angle measure defined by arc length divided by radius. The constant π is the ratio of a circle's circumference to its diameter.

The related functions are tan x = sin x/cos x, cot x = cos x/sin x, sec x = 1/cos x and cosec x = 1/sin x. These are tangent, cotangent, secant and cosecant respectively, defined when their denominators are non-zero.

Each formula applies where the function is defined. Notation such as sin² x means (sin x)². In a product, differentiating both factors and multiplying the results is incorrect: the product rule contains two separate terms.

How does the chain rule handle composite functions?

Theorem: Chain rule

Let t = u(x) be an intermediate variable and y = v(t). If the required derivatives exist, then dy/dx = (dy/dt)(dt/dx). Differentiate the outer function with its input unchanged, then multiply by the derivative of that input.

The order matters. First identify the innermost expression, then identify the function applied to it. The final answer should normally be expressed in the original variable. For several nested functions, continue multiplying by each inner derivative until reaching x.

Worked example 5. Differentiate f(x) = (2x + 1)³.

Answer: Put t = 2x + 1. The outer derivative is 3t² and dt/dx = 2. Hence f′(x) = 3t² × 2 = 6(2x + 1)². Expanding the cube first gives the same derivative.

Worked example 6. Differentiate f(x) = sin(x²).

Answer: Put t = x². Differentiating sin t gives cos t, while dt/dx = 2x. Therefore f′(x) = 2x cos(x²). The factor 2x comes from differentiating the inner square.

How do we differentiate with respect to another function?

When u and v both depend on x, differentiation of u with respect to v uses du/dv = (du/dx)/(dv/dx), provided dv/dx ≠ 0. Both derivatives in the ratio must be taken with respect to the same variable.

For sin(x³) with respect to x³, put v = x³. The function becomes sin v, whose derivative with respect to v is cos v. Thus the answer is cos(x³). This differs from differentiating with respect to x, which gives 3x² cos(x³).

The direct substitution also avoids dividing by 3x² at zero. The ratio method requires a non-zero denominator; a simplified answer cannot by itself justify a division at a point where that condition fails.

How are implicitly defined functions differentiated?

An explicit function is written as y = f(x), isolating the dependent variable y, whose value depends on the independent input x. An implicit relation connects x and y without necessarily isolating y. Implicit differentiation treats y as a differentiable function of x on the part of the relation being considered.

For example, x − y = π can be rearranged as y = x − π. The relation x + sin(xy) − y = 0 does not seem to have an easy rearrangement giving y explicitly. Direct differentiation can still produce a relation involving dy/dx.

Which factors must be retained?

1. Differentiate every term on both sides with respect to x.
2. Use the chain rule whenever a term contains y. Thus d(sin y)/dx = cos y × dy/dx.
3. Apply the product rule to mixed terms such as xy, whose derivative is y + x dy/dx.
4. Collect all terms containing dy/dx and solve, retaining any non-zero denominator condition.

Worked example 7. Find dy/dx from y + sin y = cos x, where y is differentiable and 1 + cos y ≠ 0.

Answer: Differentiation gives dy/dx + cos y × dy/dx = −sin x. Factoring gives (1 + cos y)dy/dx = −sin x. Therefore dy/dx = −sin x/(1 + cos y).

The answer can contain both x and y. There is no requirement to solve the original relation explicitly before differentiating it. However, a vanishing denominator prevents use of the displayed quotient at that point; return to the differentiated relation to investigate it.

A common error is to replace d(sin y)/dx by cos y alone. That is the derivative with respect to y. Since the requested variable is x, the inner derivative dy/dx is essential. The same distinction applies to powers and exponentials involving y.

How are inverse trigonometric functions differentiated?

An inverse trigonometric function returns an angle from a trigonometric value, using a chosen principal range to make the answer single-valued. The notation sin⁻¹ x means inverse sine, also called arcsine. It does not mean 1/sin x, which is cosec x.

For y = sin⁻¹ x, the principal range is [−π/2, π/2]. Rewrite as x = sin y. Differentiating gives 1 = cos y × dy/dx. When −1 < x < 1, cos y is positive and equals √(1 − x²), where √ means the non-negative square root.

What are the standard inverse derivatives?

The principal ranges used here are [0, π] for inverse cosine, (−π/2, π/2) for inverse tangent and (0, π) for inverse cotangent. For inverse secant use [0, π] excluding π/2; for inverse cosecant use [−π/2, π/2] excluding zero.

For a composite inverse function, replace x in the appropriate formula by its inner expression and multiply by the inner derivative. The domain of the derivative can be smaller than the domain of the function: inverse sine has endpoint values, but no finite derivative at ±1.

How can substitution simplify an inverse expression?

Worked example 8. Differentiate y = cos⁻¹[(1 − x²)/(1 + x²)] for 0 < x < 1.

Answer: Set x = tan θ, where θ is an angle in (0, π/4). The inner fraction becomes cos 2θ. Since 2θ lies in the principal range of inverse cosine, y = 2θ = 2tan⁻¹ x. Therefore dy/dx = 2/(1 + x²).

Substitution must respect the principal range. An inverse function reverses the corresponding trigonometric function on its chosen range. Cancelling the two names without checking the angle can give the wrong expression and the wrong derivative.

What are exponential and logarithmic functions?

An exponential function has a fixed positive base and a variable exponent. For b > 1, y = bˣ has every real input and positive outputs. It passes through (0, 1), increases with x and approaches zero for increasingly negative inputs.

The logarithm reverses exponentiation: logᵦ x = y means bʸ = x. Here b is the base, x > 0 is the argument and y is the logarithm. For b > 1, the logarithmic function has positive inputs and every real number as a possible output.

The constant e, between 2 and 3, is the base of the natural exponential function eˣ. Its inverse is the natural logarithm. Often the natural logarithm is denoted by ln; here it is written ln x. Logarithms to base 10 are common logarithms. All unqualified logarithms in the formulas below are written explicitly as ln.

Which logarithm laws help differentiation?

For positive quantities u and v, ln(uv) = ln u + ln v and ln(u/v) = ln u − ln v. For a real constant r and u > 0, ln(uʳ) = r ln u. These laws turn products into sums and powers into factors.

The change of base rule gives logᵦ x = ln x/ln b for b > 0, b ≠ 1 and x > 0. The base is fixed when differentiating. Do not apply logarithm laws to non-positive arguments in real-valued work.

Worked example 9. Differentiate y = sin(ln x), where x > 0.

Answer: The outer derivative is cos(ln x), and the inner derivative is 1/x. The chain rule therefore gives dy/dx = cos(ln x)/x. The positive-input condition comes from the logarithm.

For a differentiable positive function u, the composite logarithm has derivative u′/u. Similarly, the derivative of e raised to u is e raised to u multiplied by u′. Both formulas retain the inner derivative.

When is logarithmic differentiation useful?

Logarithmic differentiation means taking logarithms before differentiating. It is useful when both the base and exponent vary, or when a product, quotient or root becomes easier after applying logarithm laws. A variable exponent cannot be treated as a constant power.

Let u = u(x) be positive and v = v(x) be differentiable. For y = uᵛ, taking natural logarithms gives ln y = v ln u. Differentiating uses both the chain rule on the left and the product rule on the right.

The resulting formula is y′ = y[v′ ln u + vu′/u]. Both contributions matter: one comes from the changing exponent and the other from the changing base. Here y and u must be positive so that their real logarithms are defined.

Worked example 10. Differentiate y = x raised to sin x, for x > 0.

Answer: ln y = sin x ln x. Hence y′/y = cos x ln x + sin x/x. Multiplying by y gives $y'=x^{\sin x}(\cos x\ln x+\sin x/x)$.

How are repeated powers handled?

For y = xˣ with x > 0, write ln y = x ln x. The product rule gives y′/y = ln x + 1, so y′ = xˣ(1 + ln x). The additional logarithm term records the changing exponent.

A power tower means repeated exponentiation with the upper expression acting as the exponent of the lower base. For a finite tower y = xᵛ, let v denote the remaining upper tower. Then y′/y = v′ ln x + v/x, and differentiate v in the same way.

For an infinite repeated tower, replacing its upper tower by y gives y = xʸ only when successive finite towers converge, meaning they approach a finite positive value. On a differentiable branch, meaning a solution y that varies differentiably with x, ln y = y ln x gives y′ = y²/[x(1 − y ln x)], provided the denominator is non-zero.

Note: The repeated-power calculation assumes convergence and differentiability of the infinite tower. It does not prove either assumption. Keep these conditions attached to the formula instead of applying it to every positive x.

How do parametric equations give a derivative?

A parameter is a third variable used to express two variables separately. In a parametric description x = f(t), y = g(t), t is the parameter. The two equations together describe the relationship between x and y.

The chain rule gives dy/dt = (dy/dx)(dx/dt). Therefore dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0. Differentiate both equations with respect to the same parameter before forming this ratio.

How should the calculation be organised?

1. Identify the parameter and all constants in the two given equations.
2. Compute dx/dt from the equation defining x.
3. Compute dy/dt from the equation defining y.
4. Divide dy/dt by dx/dt and state where the denominator is non-zero.

Worked example 11. Find dy/dx if x = at² and y = 2at, where a is a non-zero constant and t ≠ 0.

Answer: dx/dt = 2at and dy/dt = 2a. Their ratio is dy/dx = 2a/(2at) = 1/t. The assumptions ensure that the denominator used in this calculation is non-zero.

Worked example 12. Find dy/dx if x = a cos θ and y = a sin θ, where a is a non-zero constant and sin θ ≠ 0.

Answer: The parameter is the angle θ. We obtain dx/dθ = −a sin θ and dy/dθ = a cos θ. Thus dy/dx = −cos θ/sin θ = −cot θ.

The derivative can remain expressed in terms of the parameter. Eliminating the parameter first is unnecessary for this method. If dx/dt = 0, the quotient rule for parametric differentiation cannot directly supply a finite derivative there, even if algebraic cancellation looks possible nearby.

How are second order derivatives found?

The second order derivative is the derivative of the first derivative, provided the first derivative is itself differentiable. For y = f(x), write it as f″(x), y″ or d²y/dx². These notations mean differentiating twice, not squaring dy/dx.

First find dy/dx and simplify it if that helps. Then differentiate every term again using the appropriate rules. A chain rule factor can appear at both stages, while a product created by the first differentiation requires the product rule at the second stage.

Worked example 13. Find the second derivative of y = x³ + tan x where cos x ≠ 0.

Answer: First, y′ = 3x² + sec² x. Next, y″ = 6x + 2sec x(sec x tan x) = 6x + 2sec² x tan x. Differentiating sec² x requires the chain rule.

How can a differential identity be verified?

A differential identity is a relation involving a function and its derivatives that holds throughout the stated domain. Compute the required derivatives, substitute them into the proposed relation and simplify. Keep the original expression for y available during substitution.

Worked example 14. If y = A sin x + B cos x, where A and B are constants, prove y″ + y = 0.

Answer: y′ = A cos x − B sin x. Differentiating again gives y″ = −A sin x − B cos x = −y. Therefore y″ + y = 0 for every real x.

For a parametric curve, first express dy/dx in terms of t. Then apply the chain rule again: d²y/dx² = [d(dy/dx)/dt]/(dx/dt), when the derivatives exist and dx/dt ≠ 0. Merely differentiating dy/dx with respect to t gives a different derivative.

A complete calculation identifies the variable at each stage. The expressions dy/dt, dy/dx and d²y/dx² answer different questions, even though they can be connected through the chain rule.

Glossary

• Continuity — Equality between a function's limiting value and its assigned value at the point being considered.
• One-sided limit — Value approached by a function when its input approaches a point from one specified side.
• Removable discontinuity — A defect repaired by assigning the existing finite limit as the value at the affected input.
• Differentiability — Existence of a finite derivative, requiring equal one-sided derivatives at an interior point.
• Difference quotient — Ratio of the change in a function's value to the corresponding non-zero change in its input.
• Modulus — The non-negative absolute value of a real number, equal to its distance from zero.
• Greatest integer function — Function assigning to each real input the largest integer less than or equal to that input.
• Composite function — A function obtained by applying one function to the output of another function.
• Chain rule — Rule differentiating a composite function by multiplying the outer derivative by the inner derivative.
• Implicit differentiation — Differentiation of a relation while treating the dependent variable as a function of the independent variable.
• Principal range — Chosen interval of angles that makes an inverse trigonometric function return a single value.
• Natural logarithm — Logarithm to base e, defined for positive real arguments and denoted by ln.
• Logarithmic differentiation — Method taking logarithms before differentiation to simplify variable powers, products and quotients.
• Parameter — A third variable through which two other variables are separately expressed in a relation.
• Second order derivative — Derivative of the first derivative, defined wherever that first derivative is itself differentiable.

Common errors and misconceptions

• Misconception: Equal one-sided limits prove continuity without any further check. Correct: Their common finite value must also equal the actual function value at the point.
• Misconception: Every continuous function is differentiable. Correct: |x| is continuous at zero but has one-sided derivatives −1 and 1, so it is not differentiable there.
• Misconception: The greatest integer function has derivative zero everywhere. Correct: Its derivative is zero between integers, but it is discontinuous and not differentiable at integers.
• Misconception: A composite derivative needs only the outer derivative. Correct: Multiply by the inner derivative; for sin(x²), the result is 2x cos(x²).
• Misconception: sin⁻¹ x is the reciprocal of sin x. Correct: It is inverse sine on its principal range; the reciprocal is cosec x.
• Misconception: The constant-power rule differentiates xˣ directly. Correct: Both base and exponent vary. Logarithmic differentiation gives xˣ(1 + ln x), for x > 0.
• Misconception: Parametric differentiation gives dx/dt divided by dy/dt. Correct: For dy/dx, divide dy/dt by dx/dt, requiring the latter to be non-zero.
• Misconception: The second derivative is the square of the first derivative. Correct: Differentiate the first derivative again with respect to the stated independent variable.

Exam-style questions with model answers

Q1. Define continuity at an interior point c of the domain of a real function f. State the one-sided limit test. [2 marks]

1. Continuity at c means that f(c) is defined and the limit of f(x) as x approaches c equals f(c).
2. The left hand limit, right hand limit and f(c) must all exist as finite real numbers and be equal.

Q2. For f(x) = x³ + 3 when x ≠ 0 and f(0) = 1, test continuity at zero and explain how to remove the discontinuity. [3 marks]

1. The assigned value is f(0) = 1. This comes from the separate definition at zero, so it must not be replaced by substituting into the other branch.
2. For non-zero inputs, f(x) = x³ + 3. Its limit as x approaches zero from either side is 3.
3. Since 3 ≠ 1, the function is discontinuous at zero. Redefining f(0) = 3 makes the limit and function value agree, removing the discontinuity.

Q3. Show that f(x) = |x| is continuous but not differentiable at x = 0, using the definition of the derivative. [5 marks]

1. By the definition of modulus, f(x) = −x for x < 0 and f(x) = x for x ≥ 0. In particular, f(0) = 0.
2. The left hand limit is the limit of −x, namely zero. The right hand limit is the limit of x, also zero. Both equal f(0), proving continuity.
3. For a negative non-zero increment h, the difference quotient [f(h) − f(0)]/h equals (−h)/h = −1. Therefore the left hand derivative is −1.
4. For a positive increment h, the difference quotient equals h/h = 1. Therefore the right hand derivative is 1.
5. The one-sided derivatives are finite but unequal. Hence the derivative at zero does not exist, proving that continuity does not imply differentiability.

Q4. Differentiate f(x) = sin(x²) with respect to x, showing how the chain rule applies. [3 marks]

1. Set t = x², so the function becomes f = sin t. This identifies the square as the inner function and sine as the outer function.
2. The derivative of the outer function with respect to t is cos t, while the derivative of the inner function with respect to x is 2x.
3. By the chain rule, f′(x) = cos t × 2x = 2x cos(x²). Both functions are differentiable for every real x.

Q5. Given y + sin y = cos x, with y differentiable as a function of x and 1 + cos y ≠ 0, find dy/dx. [3 marks]

1. Differentiate both sides with respect to x. The derivative of y is dy/dx, and the chain rule makes the derivative of sin y equal to cos y × dy/dx.
2. The resulting equation is dy/dx + cos y × dy/dx = −sin x. Collecting the two derivative terms gives (1 + cos y)dy/dx = −sin x.
3. Divide by the stated non-zero factor to obtain dy/dx = −sin x/(1 + cos y). The implicit answer can contain both x and y.

Q6. Differentiate y = cos⁻¹[(1 − x²)/(1 + x²)] for 0 < x < 1 by a trigonometric substitution. [5 marks]

1. Put x = tan θ. The given interval 0 < x < 1 allows θ to be chosen in (0, π/4), so the substitution has a definite principal angle.
2. Substitute into the inner expression: (1 − x²)/(1 + x²) = (1 − tan² θ)/(1 + tan² θ) = cos 2θ.
3. Since 0 < 2θ < π/2, the angle 2θ lies in the principal range [0, π] of inverse cosine. Thus cos⁻¹(cos 2θ) = 2θ.
4. Consequently y = 2θ = 2tan⁻¹ x. This expresses the original function in a simpler form throughout the given interval.
5. Differentiating gives dy/dx = 2/(1 + x²), for 0 < x < 1. The range check justifies the simplification used before differentiation.

Q7. For x > 0, find the derivative of y = xˣ by logarithmic differentiation. [4 marks]

1. Since x > 0, the function y = xˣ is positive, so taking real natural logarithms is valid.
2. Taking logarithms gives ln y = x ln x. The variable exponent becomes a factor, producing a product on the right.
3. Differentiation gives y′/y = ln x + x(1/x) = ln x + 1, using the chain rule and product rule.
4. Multiplying by y and replacing y by xˣ gives y′ = xˣ(1 + ln x), valid for positive x.

Q8. Let x = at² and y = 2at, where a is a fixed non-zero constant and t ≠ 0. Find dy/dx without eliminating t. [3 marks]

1. Differentiate x = at² with respect to the parameter t. Since a is constant, dx/dt = 2at, which is non-zero under the given assumptions.
2. Differentiate y = 2at with respect to the same parameter. This gives dy/dt = 2a because the derivative of t with respect to itself is 1.
3. Use dy/dx = (dy/dt)/(dx/dt) = 2a/(2at) = 1/t. The result is expressed in terms of the parameter, as permitted.

Q9. If y = A sin x + B cos x, where A and B are fixed real constants, find y′ and y″ and prove y″ + y = 0. [3 marks]

1. Differentiate term by term, keeping A and B fixed. Since the derivatives of sine and cosine are cosine and negative sine, y′ = A cos x − B sin x.
2. Differentiate again with respect to x. This gives y″ = −A sin x − B cos x, which is the negative of the given function y.
3. Substitute into the required expression: y″ + y = (−A sin x − B cos x) + (A sin x + B cos x) = 0.

Key takeaways

• Continuity at an interior point requires equal left and right hand limits, both matching the assigned function value.
• A removable discontinuity has a finite common limit; define or redefine the affected value to match that limit.
• Differentiability implies continuity, but continuity alone does not guarantee a derivative, as the modulus function shows at zero.
• Use the chain rule for composite functions, retaining every inner derivative and checking the conditions of the outer formula.
• Implicit differentiation treats y as a function of x, so differentiating a function of y introduces dy/dx.
• Inverse trigonometric simplification requires principal-range checks; the function's domain and its derivative's domain need not coincide.
• Logarithmic differentiation handles variable bases and exponents by turning a power into a product of functions.
• Parametric differentiation divides derivatives with respect to the same parameter; second differentiation requires another derivative with respect to x.

Test yourself

Can equal one-sided limits alone establish continuity at c?

No. The common finite limit must also equal the defined value f(c).

For a function defined on [a, b], which limits test continuity at its endpoints?

Use the right hand limit at a and the left hand limit at b, comparing each with its endpoint value.

Why can a jump discontinuity not be removed by changing one function value?

The left and right hand limits differ. Changing the value at the point cannot make those limits agree.

Where is the greatest integer function [x] differentiable?

It is differentiable at every non-integer input, with derivative zero, and is not differentiable at integers.

What is the derivative of e raised to cos x?

The chain rule gives −sin x multiplied by e raised to cos x.

Why must x be positive when logarithmically differentiating xˣ using ln x?

The real natural logarithm ln x is defined only for positive arguments.

What condition is needed for dy/dx = (dy/dt)/(dx/dt)?

The required parameter derivatives must exist, and the denominator dx/dt must be non-zero.

What does d²y/dx² mean, and how is it different from (dy/dx)²?

It means differentiating dy/dx again with respect to x. The other expression squares the first derivative.