Question 1
Evaluate lim x→3 (x+3).
Steps only
A polynomial is continuous, so substitute x=3.
3+3=6.
Answer
6
CLASS 11 · MATHEMATICS · CHAPTER 12
Limits and Derivatives
Begin calculus by evaluating limits of algebraic and trigonometric expressions. Short working is shown first, with a plain-language explanation for each limit.
94 worked answersAll 2 exercise setsAll 30 end-of-chapter questions
What you will learn
- Evaluate limits by direct substitution when the function is continuous.
- Simplify indeterminate 0/0 forms by factoring or rewriting.
- Use standard trigonometric limits near zero.
- Compare one-sided limits for piecewise functions.
- Find derivatives from first principles and apply derivative rules.
94 worked answers · short steps first, explanations when you need them
Exercise 12.1 — Limits
Textbook pages 237–238
Question 2
Evaluate lim x→π (x−22/7).
Steps only
Substitute x=π.
π−22/7.
Answer
π−22/7
Question 3
Evaluate lim r→1 πr².
Steps only
Substitute r=1.
π(1)²=π.
Answer
π
Question 4
Evaluate lim x→4 (4x+3)/(x−2).
Steps only
The denominator at 4 is 2, not zero; substitute directly.
(4·4+3)/(4−2)=19/2.
Answer
19/2
Question 5
Evaluate lim x→−1 (x¹⁰+x⁵+1)/(x−1).
Steps only
Substitute x=−1; the denominator is −2.
(1−1+1)/(−2)=−1/2.
Answer
−1/2
Question 6
Evaluate lim x→0 ((x+1)⁵−1)/x.
Steps only
Let u=x+1; then u→1 and x=u−1. Use the difference-of-powers limit.
lim (u⁵−1)/(u−1)=5·1⁴=5.
Answer
5
Question 7
Evaluate lim x→2 (3x²−x−10)/(x²−4).
Steps only
Factor both polynomials and cancel the common factor x−2.
((x−2)(3x+5))/((x−2)(x+2)) → 11/4.
Answer
11/4
Question 8
Evaluate lim x→3 (x⁴−81)/(2x²−5x−3).
Steps only
Factor numerator as (x−3)(x+3)(x²+9), denominator as (x−3)(2x+1); cancel x−3.
(6·18)/7=108/7.
Answer
108/7
Question 9
Evaluate lim x→0 (ax+b)/(cx+1).
Steps only
Substitute x=0; the denominator becomes 1.
(0+b)/(0+1)=b.
Answer
b
Question 10
Evaluate lim z→1 (z^(1/3)−1)/(z^(1/6)−1).
Steps only
Divide numerator and denominator by z−1 and use the power-difference limit at 1.
(1/3)/(1/6)=2.
Answer
2
Question 11
Evaluate lim x→1 (ax²+bx+c)/(cx²+bx+a), where a+b+c≠0.
Steps only
Substitute x=1. Both numerator and denominator become a+b+c, which is nonzero.
(a+b+c)/(a+b+c)=1.
Answer
1
Question 12
Evaluate lim x→−2 (1/x+1/2)/(x+2).
Steps only
Combine the terms in the numerator: 1/x+1/2=(x+2)/(2x); cancel x+2.
1/(2x) → −1/4.
Answer
−1/4
Question 13
Evaluate lim x→0 sin(ax)/(bx), for constants a,b.
Steps only
Rewrite as (a/b)·sin(ax)/(ax).
(a/b)·1=a/b.
Answer
a/b
Question 14
Evaluate lim x→0 sin(ax)/sin(bx), where a,b≠0.
Steps only
Insert ax and bx to form standard sine ratios.
[sin(ax)/(ax)]/[sin(bx)/(bx)]·a/b → a/b.
Answer
a/b
Question 15
Evaluate lim x→π sin(π−x)/[π(π−x)].
Steps only
Set u=π−x. Then u→0.
(1/π)·lim sin u/u=1/π.
Answer
1/π
Question 16
Evaluate lim x→0 cos x/(π−x).
Steps only
Substitute x=0; the denominator is π.
cos 0/π=1/π.
Answer
1/π
Question 17
Evaluate lim x→0 (cos 2x−1)/(cos x−1).
Steps only
Use cos 2x−1=−2sin²x and cos x−1=−2sin²(x/2); take the ratio.
sin²x/sin²(x/2) → (x/(x/2))²=4.
Answer
4
Question 18
Evaluate lim x→0 (ax+x cos x)/(b sin x).
Steps only
Factor x from the numerator and denominator, then form sin x/x.
(a+cos x)/(b·sin x/x) → (a+1)/b.
Answer
(a+1)/b
Question 19
Evaluate lim x→0 x sec x.
Steps only
Write sec x=1/cos x and substitute.
0/cos 0=0.
Answer
0
Question 20
Evaluate lim x→0 (sin ax+bx)/(ax+sin bx), where a+b≠0.
Steps only
Divide top and bottom by x and use sin(cx)/(cx)→1.
(a+b)/(a+b)=1.
Answer
1
Question 21
Evaluate lim x→0 (cosec x−cot x).
Steps only
Combine over a common denominator: (1−cos x)/sin x.
[(1−cos x)/x]/[sin x/x] → 0/1=0.
Answer
0
Question 22
Evaluate lim x→π/2 tan(2x)/(x−π/2).
Steps only
Set u=x−π/2. Then tan(2x)=tan(2u+π)=tan(2u).
[tan(2u)/(2u)]·2 → 2.
Answer
2
Question 23(i)
For f(x)=2x+3 when x≤0 and f(x)=3(x+1) when x>0, find lim x→0 f(x).
Steps only
Approach 0 from each side using the matching rule.
LHL=2(0)+3=3; RHL=3(0+1)=3.
Answer
3.
Question 23(ii)
For the same piecewise function, find lim x→1 f(x).
Steps only
Values near 1 use the branch f(x)=3(x+1).
3(1+1)=6.
Answer
6.
Question 24
For f(x)=x²−1 when x≤1 and f(x)=−x²−1 when x>1, find lim x→1 f(x).
Steps only
Calculate the left-hand and right-hand limits separately.
LHL=1²−1=0; RHL=−1²−1=−2.
Answer
The limit does not exist because 0≠−2.
Question 25
For f(x)=|x|/x when x≠0 and f(0)=0, evaluate lim x→0 f(x).
Steps only
For x<0, |x|/x=−1; for x>0, |x|/x=1.
LHL=−1 and RHL=1.
Answer
The limit does not exist.
Question 26
For f(x)=x/|x| when x≠0 and f(0)=0, find lim x→0 f(x).
Steps only
For positive x the quotient is 1; for negative x it is −1.
LHL=−1; RHL=1.
Answer
The limit does not exist.
Question 27
Find lim x→5 (|x|−5).
Steps only
Near 5, x is positive, so |x|=x.
x−5 → 5−5=0.
Answer
0.
Question 28
If f(x)=a+bx for x<1, f(1)=4, and f(x)=b−ax for x>1, find a and b given lim x→1 f(x)=f(1).
Steps only
Set each one-sided limit equal to 4.
a+b=4; b−a=4. Adding and subtracting gives b=4 and a=0.
Answer
a=0, b=4.
Question 29(i)
For f(x)=(x−a₁)(x−a₂)…(x−aₙ), find lim x→a₁ f(x).
Steps only
Substitute a₁ into the polynomial.
(a₁−a₁)(a₁−a₂)…(a₁−aₙ)=0.
Answer
0.
Question 29(ii)
For f(x)=(x−a₁)(x−a₂)…(x−aₙ), find lim x→a f(x) when a differs from every aᵢ.
Steps only
Substitute x=a into the polynomial.
(a−a₁)(a−a₂)…(a−aₙ).
Answer
(a−a₁)(a−a₂)…(a−aₙ).
Question 30
If f(x)=|x|+1 for x<0, f(0)=0, and f(x)=x−1 for x>0, find where lim x→a f(x) exists.
Steps only
At a=0 compare the sides; at any nonzero a, one continuous formula applies nearby.
At 0: LHL=1, RHL=−1. For a≠0 the limit exists.
Answer
The limit exists for every a≠0 and does not exist at a=0.
Question 31
If lim x→1 [f(x)−2]/(x²−1)=π, find lim x→1 f(x).
Steps only
The denominator approaches zero while the quotient has a finite limit.
Therefore f(x)−2→0, so f(x)→2.
Answer
2.
Question 32
If f(x)=mx²+n for x<0, nx+m for 0≤x≤1, and nx³+m for x>1, find when both limits at 0 and 1 exist.
Steps only
At zero compare n with m. At one compare the middle and right formulas.
At 0: LHL=n, RHL=m, so m=n. At 1 both sides equal n+m.
Answer
The limit at 0 exists exactly when m=n; the limit at 1 exists for all integers m,n.
Exercise 12.2 — Derivatives
Textbook pages 249
Question 1
Find the derivative of x²−2 at x=10.
Steps only
f(x)=x²−2.
f′(x)=2x, so f′(10)=20.
Answer
20
Question 2
Find the derivative of x at x=1.
Steps only
The derivative of x is 1.
f′(1)=1.
Answer
1
Question 3
Find the derivative of 99x at x=100.
Steps only
The derivative of 99x is 99.
f′(100)=99.
Answer
99
Question 4(i)
Use first principles to differentiate x³−27.
Steps only
f′(x)=limₕ→₀[((x+h)³−27)−(x³−27)]/h.
Expand, cancel h: [3x²h+3xh²+h³]/h → 3x².
Answer
3x²
Question 4(ii)
Use first principles to differentiate (x−1)(x−2).
Steps only
First expand f(x)=x²−3x+2.
[f(x+h)−f(x)]/h=2x+h−3 → 2x−3.
Answer
2x−3
Question 4(iii)
Use first principles to differentiate 1/x².
Steps only
f′(x)=limₕ→₀[1/(x+h)²−1/x²]/h.
Combine fractions: −(2x+h)/[x²(x+h)²] → −2/x³.
Answer
−2/x³, for x≠0
Question 4(iv)
Use first principles to differentiate (x+1)/(x−1).
Steps only
Set f(x)=(x+1)/(x−1) and form [f(x+h)−f(x)]/h.
The numerator after combining fractions is −2h; cancel h to get −2/[(x+h−1)(x−1)] → −2/(x−1)².
Answer
−2/(x−1)², for x≠1
Question 5
For f(x)=x¹⁰⁰/100+x⁹⁹/99+⋯+x²/2+x+1, prove f′(1)=100f′(0).
Steps only
Differentiate each term: f′(x)=x⁹⁹+x⁹⁸+⋯+x+1.
f′(1)=100; f′(0)=1; hence f′(1)=100f′(0).
Answer
Proved: f′(1)=100f′(0).
Question 6
Differentiate xⁿ+axⁿ⁻¹+a²xⁿ⁻²+⋯+aⁿ⁻¹x+aⁿ, where a is constant.
Steps only
Differentiate each power of x; a is fixed.
f′(x)=nxⁿ⁻¹+a(n−1)xⁿ⁻²+a²(n−2)xⁿ⁻³+⋯+aⁿ⁻¹.
Answer
nxⁿ⁻¹+a(n−1)xⁿ⁻²+a²(n−2)xⁿ⁻³+⋯+aⁿ⁻¹.
Question 7(i)
Differentiate (x−a)(x−b), where a and b are constants.
Steps only
Expand: x²−(a+b)x+ab.
Differentiate termwise: 2x−(a+b).
Answer
2x−a−b.
Question 7(ii)
Differentiate (ax²+b)², where a and b are constants.
Steps only
Use the chain rule: 2(ax²+b)·(2ax).
f′(x)=4ax(ax²+b).
Answer
4ax(ax²+b).
Question 7(iii)
Differentiate (x−a)/(x−b), where a and b are constants.
Steps only
Use the quotient rule with numerator and denominator derivatives both 1.
[(x−b)−(x−a)]/(x−b)²=(a−b)/(x−b)².
Answer
(a−b)/(x−b)², for x≠b.
Question 8
Differentiate (xⁿ−aⁿ)/(x−a), where a is constant.
Steps only
Apply the quotient rule: u′=nxⁿ⁻¹ and v′=1.
f′(x)=[nxⁿ⁻¹(x−a)−(xⁿ−aⁿ)]/(x−a)².
Answer
[nxⁿ⁻¹(x−a)−xⁿ+aⁿ]/(x−a)², for x≠a.
Question 9(i)
Differentiate 2x−3/4.
Steps only
The derivative of 2x is 2; the constant has derivative zero.
f′(x)=2.
Answer
2.
Question 9(ii)
Differentiate (5x³+3x−1)(x−1).
Steps only
Expand: 5x⁴−5x³+3x²−4x+1.
Differentiate termwise: 20x³−15x²+6x−4.
Answer
20x³−15x²+6x−4.
Question 9(iii)
Differentiate x⁻³(5+3x).
Steps only
Rewrite as 5x⁻³+3x⁻².
f′(x)=−15x⁻⁴−6x⁻³=−3(5+2x)/x⁴.
Answer
−3(5+2x)/x⁴, for x≠0.
Question 9(iv)
Differentiate x⁵(3−6x⁻⁹).
Steps only
Expand: 3x⁵−6x⁻⁴.
f′(x)=15x⁴+24x⁻⁵.
Answer
15x⁴+24/x⁵, for x≠0.
Question 9(v)
Differentiate x⁻⁴(3−4x⁻⁵).
Steps only
Expand: 3x⁻⁴−4x⁻⁹.
f′(x)=−12x⁻⁵+36x⁻¹⁰.
Answer
−12/x⁵+36/x¹⁰, for x≠0.
Question 9(vi)
Differentiate 2/(x+1)−x²/(3x−1).
Steps only
Differentiate the first quotient and use the quotient rule on the second.
−2/(x+1)²−[2x(3x−1)−3x²]/(3x−1)².
Answer
−2/(x+1)²−x(3x−2)/(3x−1)², for x≠−1,1/3.
Question 10
Use first principles to differentiate cos x.
Steps only
f′(x)=limₕ→₀[cos(x+h)−cos x]/h. Use the cosine-difference identity.
=−sin(x+h/2)·[sin(h/2)/(h/2)] → −sin x.
Answer
−sin x.
Question 11(i)
Differentiate sin x cos x.
Steps only
Use the product rule: (sin x)′cos x+sin x(cos x)′.
cos²x−sin²x=cos 2x.
Answer
cos 2x.
Question 11(ii)
Differentiate sec x.
Steps only
Write sec x=1/cos x and differentiate the quotient.
sin x/cos²x=sec x tan x.
Answer
sec x tan x.
Question 11(iii)
Differentiate 5 sec x+4 cos x.
Steps only
Use the sum rule and constant multiples.
5 sec x tan x−4 sin x.
Answer
5 sec x tan x−4 sin x.
Question 11(iv)
Differentiate cosec x.
Steps only
Write cosec x=1/sin x and differentiate.
−cos x/sin²x=−cosec x cot x.
Answer
−cosec x cot x.
Question 11(v)
Differentiate 3 cot x+5 cosec x.
Steps only
Apply the sum rule and constant multiples.
−3 cosec²x−5 cosec x cot x.
Answer
−3 cosec²x−5 cosec x cot x.
Question 11(vi)
Differentiate 5 sin x−6 cos x+7.
Steps only
Differentiate each term separately.
5 cos x+6 sin x+0.
Answer
5 cos x+6 sin x.
Question 11(vii)
Differentiate 2 tan x−7 sec x.
Steps only
Apply the sum rule and constant multiples.
2 sec²x−7 sec x tan x.
Answer
2 sec²x−7 sec x tan x.
Miscellaneous Exercise on Chapter 12
Textbook pages 254–255
Question 1(i)
Differentiate −x from first principles.
Steps only
f′(x)=limₕ→₀[−(x+h)+x]/h.
=limₕ→₀(−h)/h=−1.
Answer
−1.
Question 1(ii)
Differentiate 1/(−x) from first principles.
Steps only
Write f(x)=−1/x and use [f(x+h)−f(x)]/h.
The numerator simplifies to h/[x(x+h)]; divide by h and let h→0.
Answer
1/x², for x≠0.
Question 1(iii)
Differentiate sin(x+1) from first principles.
Steps only
Use f′(x)=limₕ→₀[sin(x+1+h)−sin(x+1)]/h and the sine-sum identity.
=lim 2cos(x+1+h/2)·sin(h/2)/h=cos(x+1).
Answer
cos(x+1).
Question 1(iv)
Differentiate cos(x−π/8) from first principles.
Steps only
Use the cosine difference quotient with angle x−π/8.
The cosine-difference identity and sin(h/2)/(h/2)→1 give −sin(x−π/8).
Answer
−sin(x−π/8).
Question 2
Differentiate x+a, where a is constant.
Steps only
The derivative of x is 1 and the derivative of fixed a is 0.
d(x+a)/dx=1+0.
Answer
1.
Question 3
Differentiate (px+q)(r/x+s), with p,q,r,s constant.
Steps only
Expand: pr+psx+qr/x+qs.
Derivative = ps−qr/x².
Answer
ps−qr/x², for x≠0.
Question 4
Differentiate (ax+b)(cx+d)², with constants a,b,c,d.
Steps only
Use the product rule; the two factor derivatives are a and 2c(cx+d).
a(cx+d)²+2c(ax+b)(cx+d).
Answer
a(cx+d)²+2c(ax+b)(cx+d).
Question 5
Differentiate (ax+b)/(cx+d), with constants a,b,c,d.
Steps only
Apply the quotient rule to the two linear expressions.
[a(cx+d)−c(ax+b)]/(cx+d)²=(ad−bc)/(cx+d)².
Answer
(ad−bc)/(cx+d)², where cx+d≠0.
Question 6
Differentiate (1+1/x)/(1−1/x).
Steps only
For x≠0, rewrite as (x+1)/(x−1), then use the quotient rule.
[(x−1)−(x+1)]/(x−1)²=−2/(x−1)².
Answer
−2/(x−1)², for x≠0,1.
Question 7
Differentiate 1/(ax²+bx+c), where a,b,c are constants.
Steps only
Treat it as a reciprocal; the denominator derivative is 2ax+b.
−(2ax+b)/(ax²+bx+c)².
Answer
−(2ax+b)/(ax²+bx+c)², where ax²+bx+c≠0.
Question 8
Differentiate (ax+b)/(px²+qx+r), with constants.
Steps only
Apply the quotient rule; numerator derivative is a and denominator derivative is 2px+q.
[a(px²+qx+r)−(ax+b)(2px+q)]/(px²+qx+r)².
Answer
(−apx²−2bpx+ar−bq)/(px²+qx+r)².
Question 9
Differentiate (px²+qx+r)/(ax+b), with constants.
Steps only
Use the quotient rule with derivatives 2px+q and a.
[(2px+q)(ax+b)−a(px²+qx+r)]/(ax+b)².
Answer
(apx²+2bpx+bq−ar)/(ax+b)².
Question 10
Differentiate a/x⁴−b/x²+cos x, where a,b are constants.
Steps only
Rewrite the powers as ax⁻⁴−bx⁻²+cos x.
−4ax⁻⁵+2bx⁻³−sin x.
Answer
−4a/x⁵+2b/x³−sin x, for x≠0.
Question 11
Differentiate 4√x−2.
Steps only
Write √x=x^(1/2), then apply the power rule.
4·(1/2)x^(−1/2)=2/√x.
Answer
2/√x, for x>0.
Question 12
Differentiate (ax+b)ⁿ, where a,b are constants.
Steps only
Use the chain rule: differentiate the outer nth power, then multiply by the inside derivative a.
na(ax+b)ⁿ⁻¹.
Answer
na(ax+b)ⁿ⁻¹.
Question 13
Differentiate (ax+b)ⁿ(cx+d)ᵐ, where constants a,b,c,d and integers m,n are fixed.
Steps only
Apply the product rule and chain rule to each factor.
na(ax+b)ⁿ⁻¹(cx+d)ᵐ+mc(ax+b)ⁿ(cx+d)ᵐ⁻¹.
Answer
(ax+b)ⁿ⁻¹(cx+d)ᵐ⁻¹[na(cx+d)+mc(ax+b)].
Question 14
Differentiate sin(x+a), where a is constant.
Steps only
Use sin(x+a)=sin x cos a+cos x sin a.
Derivative = cos x cos a−sin x sin a.
Answer
cos(x+a).
Question 15
Differentiate cosec x·cot x.
Steps only
Use the product rule with (cosec x)′=−cosec x cot x and (cot x)′=−cosec²x.
−cosec x cot²x−cosec³x.
Answer
−cosec³x−cosec x cot²x.
Question 16
Differentiate cos x/(1+sin x).
Steps only
Apply the quotient rule with denominator derivative cos x.
[−sin x(1+sin x)−cos²x]/(1+sin x)²=−1/(1+sin x).
Answer
−1/(1+sin x), where 1+sin x≠0.
Question 17
Differentiate (sin x+cos x)/(sin x−cos x).
Steps only
Use the quotient rule; simplify the numerator using squared sum and difference identities.
Numerator = −(sin x−cos x)²−(sin x+cos x)²=−2.
Answer
−2/(sin x−cos x)², where sin x≠cos x.
Question 18
Differentiate (sec x−1)/(sec x+1).
Steps only
Both numerator and denominator have derivative sec x tan x.
[sec x tan x((sec x+1)−(sec x−1))]/(sec x+1)².
Answer
2 sec x tan x/(sec x+1)².
Question 19
Differentiate (sin x)ⁿ, where n is an integer.
Steps only
Use the chain rule for the nth power of sin x.
n(sin x)ⁿ⁻¹ cos x.
Answer
n sinⁿ⁻¹x cos x.
Question 20
Differentiate (a+b sin x)/(c+d cos x), with constants a,b,c,d.
Steps only
Use the quotient rule with derivatives b cos x and −d sin x.
Numerator = b cos x(c+d cos x)+d sin x(a+b sin x).
Answer
(bc cos x+ad sin x+bd)/(c+d cos x)².
Question 21
Differentiate sin(x+a)/cos x, where a is constant.
Steps only
Apply the quotient rule; numerator derivative is cos(x+a), denominator derivative is −sin x.
Numerator = cos(x+a)cos x+sin(x+a)sin x=cos a.
Answer
cos a/cos²x, where cos x≠0.
Question 22
Differentiate x⁴(5 sin x−3 cos x).
Steps only
Use the product rule; the derivatives are 4x³ and 5 cos x+3 sin x.
4x³(5 sin x−3 cos x)+x⁴(5 cos x+3 sin x).
Answer
x³(5x cos x+3x sin x+20 sin x−12 cos x).
Question 23
Differentiate (x²+1)cos x.
Steps only
Use the product rule.
2x cos x−(x²+1)sin x.
Answer
2x cos x−(x²+1)sin x.
Question 24
Differentiate (ax²+sin x)(p+q cos x), with constants a,p,q.
Steps only
Use the product rule with derivatives 2ax+cos x and −q sin x.
(2ax+cos x)(p+q cos x)−q sin x(ax²+sin x).
Answer
(2ax+cos x)(p+q cos x)−q sin x(ax²+sin x).
Question 25
Differentiate (x+cos x)(x−tan x).
Steps only
Use the product rule; derivatives are 1−sin x and 1−sec²x=−tan²x.
(1−sin x)(x−tan x)−tan²x(x+cos x).
Answer
(1−sin x)(x−tan x)−tan²x(x+cos x).
Question 26
Differentiate (4x+5 sin x)/(3x+7 cos x).
Steps only
Apply the quotient rule; derivatives are 4+5 cos x and 3−7 sin x.
[(4+5 cos x)(3x+7 cos x)−(4x+5 sin x)(3−7 sin x)]/(3x+7 cos x)².
Answer
(35+15x cos x+28 cos x+28x sin x−15 sin x)/(3x+7 cos x)².
Question 27
Differentiate x² cos(π/4)/sin x.
Steps only
The factor cos(π/4) is constant; apply the quotient rule to x²/sin x.
cos(π/4)[2x sin x−x² cos x]/sin²x.
Answer
x cos(π/4)(2 sin x−x cos x)/sin²x.
Question 28
Differentiate x/(1+tan x).
Steps only
Apply the quotient rule; the denominator derivative is sec²x.
[(1+tan x)−x sec²x]/(1+tan x)².
Answer
(1+tan x−x sec²x)/(1+tan x)².
Question 29
Differentiate (x+sec x)(x−tan x).
Steps only
Use the product rule; derivatives are 1+sec x tan x and 1−sec²x.
(1+sec x tan x)(x−tan x)+(x+sec x)(1−sec²x).
Answer
(x+sec x)(1−sec²x)+(x−tan x)(1+sec x tan x).
Question 30
Differentiate x/sinⁿx.
Steps only
Use the quotient rule; (sinⁿx)′=n sinⁿ⁻¹x cos x.
[sinⁿx−nx sinⁿ⁻¹x cos x]/sin²ⁿx.
Answer
(sin x−nx cos x)/sinⁿ⁺¹x, where sin x≠0.