Question 1
Prove that f(x) = 5x − 3 is continuous at x = 0, x = −3 and x = 5.
Steps only
For any real c, limₓ→c f(x) = 5c − 3 = f(c).
So this holds at c = 0, −3 and 5.
Answer
Continuous at x = 0, −3 and 5.
CLASS 12 · MATHEMATICS PART I · CHAPTER 5
Continuity and differentiability
Short working is shown first. Select “Show explanation” for an easy step-by-step version.
137 worked answersAll 7 exercise setsAll 22 end-of-chapter questions
137 worked answers · short steps first, explanations when you need them
Exercise 5.1 · Questions 1–29
Textbook pages 116–118
Question 2
Examine the continuity of f(x) = 2x² − 1 at x = 3.
Steps only
f(3) = 2(3)² − 1 = 17.
limₓ→3 f(x) = 2(3)² − 1 = 17 = f(3).
Answer
f is continuous at x = 3.
Question 3(a)
Examine f(x) = x − 5 for continuity.
Steps only
For every real c, limₓ→c(x − 5) = c − 5 = f(c).
Answer
Continuous for every real x.
Question 3(b)
Examine f(x) = 1/(x − 5), where x ≠ 5, for continuity.
Steps only
The domain is ℝ except 5.
For c ≠ 5, limₓ→c 1/(x − 5) = 1/(c − 5) = f(c).
Answer
Continuous at every point of its domain (x ≠ 5).
Question 3(c)
Examine f(x) = (x² − 25)/(x + 5), where x ≠ −5, for continuity.
Steps only
For x ≠ −5, f(x) = (x − 5)(x + 5)/(x + 5) = x − 5.
Thus limₓ→c f(x) = c − 5 = f(c) for every c ≠ −5.
Answer
Continuous at every point of its domain (x ≠ −5).
Question 3(d)
Examine f(x) = |x − 5| for continuity.
Steps only
Absolute value is continuous, and x − 5 is linear.
So their composite |x − 5| is continuous for every real x.
Answer
Continuous for every real x.
Question 4
Prove that f(x) = xⁿ is continuous at x = n, where n is a positive integer.
Steps only
f(n) = nⁿ.
limₓ→n xⁿ = nⁿ = f(n).
Answer
f is continuous at x = n.
Question 5
For f(x) = x when x ≤ 1 and f(x) = 5 when x > 1, decide whether f is continuous at x = 0, 1 and 2.
Steps only
At 0 and 2, each nearby input uses the same continuous rule, so f is continuous.
At 1: LHL = 1, RHL = 5; they differ.
Answer
Continuous at x = 0 and 2; discontinuous at x = 1.
Question 6
Find all points of discontinuity of f(x) = 2x + 3 when x ≤ 2 and f(x) = 2x − 3 when x > 2.
Steps only
Both rules are linear, so only the joining point x = 2 needs checking.
f(2) = 7; right-hand limit = 1.
Answer
The only point of discontinuity is x = 2.
Question 8
For f(x) = |x|/x when x ≠ 0 and f(0) = 0, find its point(s) of discontinuity.
Steps only
For x < 0, |x|/x = −1; for x > 0, |x|/x = 1.
LHL at 0 = −1 and RHL at 0 = 1.
Answer
Discontinuous at x = 0.
Question 9
Examine the continuity of f(x) = x/|x| for x < 0 and f(x) = −1 for x ≥ 0.
Steps only
For x < 0, x/|x| = −1. For x ≥ 0, f(x) = −1.
Thus f(x) = −1 for all real x.
Answer
Continuous for every real x.
Question 10
Examine f(x) = x² + 1 when x ≥ 1 and f(x) = x + 1 when x < 1.
Steps only
At x = 1, LHL = 2 and RHL = 2 = f(1).
Each branch is continuous elsewhere.
Answer
Continuous for every real x.
Question 12
Find all points of discontinuity of f(x) = x¹⁰ − 1 for x ≤ 1 and f(x) = x² for x > 1.
Steps only
At x = 1, f(1) = 0 and the right-hand limit is 1.
The limits do not match.
Answer
Discontinuous at x = 1.
Question 13
Examine f(x) = x + 5 when x ≤ 1 and f(x) = x − 5 when x > 1.
Steps only
At x = 1, LHL = 6 and RHL = −4.
The one-sided limits differ.
Answer
Discontinuous at x = 1.
Question 7
Find all points of discontinuity of f(x) = |x| + 3 for x ≤ −3; f(x) = −2x for −3 < x < 3; f(x) = 6x + 2 for x ≥ 3.
Steps only
At x = −3: left value = right limit = 6.
At x = 3: left limit = −6, right limit = 20.
Answer
The only point of discontinuity is x = 3.
Question 11
Find all points of discontinuity of f(x) = x³ − 3 for x ≤ 2 and f(x) = x² + 1 for x > 2.
Steps only
At x = 2, f(2) = 8 − 3 = 5.
RHL = 2² + 1 = 5.
Answer
There are no points of discontinuity; f is continuous everywhere.
Question 14
Discuss continuity of f(x) = 3 for 0 ≤ x ≤ 1; f(x) = 4 for 1 < x < 3; f(x) = 5 for 3 ≤ x ≤ 10.
Steps only
The function is constant between the join points.
At x=1: left=3, right=4. At x=3: left=4, right=5.
Answer
Discontinuous at x = 1 and x = 3; continuous at all other points of [0, 10].
Question 15
Discuss f(x) = 2x for x < 0; f(x) = 0 for 0 ≤ x ≤ 1; f(x) = 4x for x > 1.
Steps only
At x=0: LHL=0=RHL=f(0).
At x=1: LHL=f(1)=0, RHL=4.
Answer
Discontinuous only at x = 1; continuous at x = 0 and elsewhere.
Question 16
Discuss f(x) = −2 for x ≤ −1; f(x) = 2x for −1 < x ≤ 1; f(x) = 2 for x > 1.
Steps only
At x=−1: both sides and f(−1) equal −2.
At x=1: both sides and f(1) equal 2.
Answer
Continuous for every real x.
Question 17
Find the relation between a and b so f(x) = ax + 1 for x ≤ 3 and f(x) = bx + 3 for x > 3 is continuous at x = 3.
Steps only
At 3: left value = 3a + 1; right limit = 3b + 3.
3a + 1 = 3b + 3 ⇒ 3a − 3b = 2.
Answer
3a − 3b = 2 (equivalently, a − b = 2/3).
Question 18
For f(x) = λ(x² − 2x) when x ≤ 0 and f(x) = 4x + 1 when x > 0, find λ values making f continuous at 0. Also check continuity at x = 1.
Steps only
At 0: f(0)=0 for every λ; RHL=1. No λ works.
Near 1, f(x)=4x+1, a polynomial.
Answer
No value of λ makes f continuous at x = 0. It is continuous at x = 1 for every λ.
Question 19
Show that g(x) = x − [x] is discontinuous at every integer, where [x] is the greatest integer not exceeding x.
Steps only
At any integer n: LHL = 1; RHL = g(n) = 0.
Answer
g is discontinuous at every integer n.
Question 20
Is f(x) = x² − sin x + 5 continuous at x = π?
Steps only
x², sin x and 5 are continuous everywhere.
Their sum and difference are continuous; therefore f is continuous at π.
Answer
Yes, f is continuous at x = π.
Question 21(a)
Discuss continuity of f(x) = sin x + cos x.
Steps only
sin x and cos x are continuous for all real x.
Their sum is continuous for all real x.
Answer
Continuous for every real x.
Question 21(b)
Discuss continuity of f(x) = sin x − cos x.
Steps only
sin x and cos x are continuous for all real x.
Their difference is continuous for all real x.
Answer
Continuous for every real x.
Question 21(c)
Discuss continuity of f(x) = sin x · cos x.
Steps only
sin x and cos x are continuous for all real x.
Their product is continuous for all real x.
Answer
Continuous for every real x.
Question 22
Discuss where cos x, cosec x, sec x and cot x are continuous.
Steps only
cos x: all real x.
cosec x and cot x: x ≠ nπ. sec x: x ≠ (2n+1)π/2, n∈ℤ.
Answer
cos x is continuous on ℝ; cosec x and cot x are continuous where sin x ≠ 0; sec x is continuous where cos x ≠ 0.
Question 23
Find all discontinuities of f(x) = sin x/x for x < 0 and f(x) = x + 1 for x ≥ 0.
Steps only
At 0: limₓ→0− sin x/x = 1; limₓ→0+ (x+1) = 1 = f(0).
Answer
There are no points of discontinuity; f is continuous on ℝ.
Question 24
Determine whether f(x) = x² sin(1/x) for x ≠ 0 and f(0)=0 is continuous.
Steps only
−x² ≤ x²sin(1/x) ≤ x².
Both bounds tend to 0, so limₓ→0 f(x)=0=f(0).
Answer
Yes, f is continuous everywhere.
Question 25
Examine f(x) = sin x − cos x for x ≠ 0 and f(0)=−1.
Steps only
limₓ→0(sin x−cos x)=0−1=−1=f(0).
Answer
Continuous for every real x.
Question 26
Find k so f(x)=k cos x/(π−2x) for x ≠ π/2 and f(π/2)=3 is continuous at x=π/2.
Steps only
Let h=x−π/2. Then cos x/(π−2x) = (sin h)/(2h) → 1/2.
k/2=3 ⇒ k=6.
Answer
k = 6.
Question 27
Find k so f(x)=kx² for x ≤ 2 and f(x)=3 for x > 2 is continuous at x=2.
Steps only
f(2)=4k; RHL=3.
4k=3 ⇒ k=3/4.
Answer
k = 3/4.
Question 28
Find k so f(x)=kx+1 for x ≤ π and f(x)=cos x for x > π is continuous at x=π.
Steps only
f(π)=kπ+1; RHL=cos π=−1.
kπ+1=−1 ⇒ k=−2/π.
Answer
k = −2/π.
Question 29
Find k so f(x)=kx+1 for x ≤ 5 and f(x)=3x−5 for x > 5 is continuous at x=5.
Steps only
f(5)=5k+1; RHL=15−5=10.
5k+1=10 ⇒ k=9/5.
Answer
k = 9/5.
Exercise 5.2
Textbook pages 122
Question 1
Differentiate y = sin(x² + 5) with respect to x.
Steps only
y′ = cos(x²+5)·2x.
Answer
2x cos(x² + 5).
Question 2
Differentiate y = cos(sin x) with respect to x.
Steps only
y′ = −sin(sin x)·cos x.
Answer
−cos x sin(sin x).
Question 3
Differentiate y = sin(ax + b), where a and b are constants.
Steps only
y′ = cos(ax+b)·a.
Answer
a cos(ax + b).
Question 4
Differentiate y = sec(tan x) with respect to x.
Steps only
y′ = sec(tan x) tan(tan x) · sec²x.
Answer
sec(tan x) tan(tan x) sec²x.
Question 5
Differentiate y = sin(ax+b)/cos(cx+d), where a,b,c,d are constants.
Steps only
Write y=sin(ax+b) sec(cx+d).
y′=a cos(ax+b)sec(cx+d)+c sin(ax+b)tan(cx+d)sec(cx+d).
Answer
a cos(ax + b) sec(cx + d) + c sin(ax + b) tan(cx + d) sec(cx + d).
Question 6
Differentiate y = cos(x³) · sin²(x⁵) with respect to x.
Steps only
y′=−3x²sin(x³)sin²(x⁵)+10x⁴cos(x³)sin(x⁵)cos(x⁵).
Answer
−3x² sin(x³) sin²(x⁵) + 10x⁴ cos(x³) sin(x⁵) cos(x⁵).
Question 7
Differentiate y = 2√(cot(x²)) with respect to x.
Steps only
y′=2·(1/2)(cot(x²))^(−1/2)·[−cosec²(x²)]·2x.
Answer
−2x cosec²(x²)/√(cot(x²)).
Question 8
Differentiate y = cos(√x) with respect to x.
Steps only
y′=−sin(√x)·1/(2√x).
Answer
−sin(√x)/(2√x), for x>0.
Question 9
Prove that f(x)=|x−1| is not differentiable at x=1.
Steps only
For x<1, f(x)=1−x, so left derivative=−1.
For x>1, f(x)=x−1, so right derivative=1.
Answer
The left and right derivatives at x=1 are unequal, so f is not differentiable there.
Question 10
Show that f(x)=[x], for 0<x<3, is not differentiable at x=1 and x=2.
Steps only
At each integer n=1,2, the left limit of [x] is n−1 and the right limit and value are n.
There is a jump, so the derivative cannot exist.
Answer
f is not differentiable at x=1 or x=2.
Exercise 5.3
Textbook pages 125
Question 1
Find dy/dx if 2x + 3y = sin x.
Steps only
2 + 3y′ = cos x.
y′ = (cos x − 2)/3.
Answer
(cos x − 2)/3.
Question 2
Find dy/dx if 2x + 3y = sin y.
Steps only
2 + 3y′ = cos y · y′.
y′ = 2/(cos y − 3).
Answer
2/(cos y − 3).
Question 3
Find dy/dx if ax + by² = cos y, where a and b are constants.
Steps only
a + 2by y′ = −sin y · y′.
y′ = −a/(2by + sin y).
Answer
−a/(2by + sin y).
Question 4
Find dy/dx if xy + y² = tan x + y.
Steps only
y + xy′ + 2yy′ = sec²x + y′.
y′(x+2y−1)=sec²x−y.
Answer
(sec²x − y)/(x + 2y − 1).
Question 5
Find dy/dx if x² + xy + y² = 100.
Steps only
2x + y + xy′ + 2yy′ = 0.
y′ = −(2x+y)/(x+2y).
Answer
−(2x + y)/(x + 2y).
Question 6
Find dy/dx if x³ + x²y + xy² + y³ = 81.
Steps only
3x² + 2xy + x²y′ + y² + 2xyy′ + 3y²y′ = 0.
y′ = −(3x²+2xy+y²)/(x²+2xy+3y²).
Answer
−(3x² + 2xy + y²)/(x² + 2xy + 3y²).
Question 7
Find dy/dx if sin²y + cos(xy) = κ, where κ is a constant.
Steps only
2sin y cos y·y′ − sin(xy)(y+xy′)=0.
y′ = y sin(xy)/(sin 2y − x sin(xy)).
Answer
y sin(xy)/(sin 2y − x sin(xy)).
Question 8
Find dy/dx if sin²x + cos²y = 1.
Steps only
2sin x cos x − 2cos y sin y·y′=0.
y′=sin 2x/sin 2y.
Answer
sin 2x/sin 2y.
Question 9
Find dy/dx if y = sin⁻¹(2x/(1+x²)). The question gives no interval for x.
Steps only
For |x|<1, y=2tan⁻¹x, so y′=2/(1+x²).
For |x|>1, y=π−2tan⁻¹x (x>1) or −π−2tan⁻¹x (x<−1), so y′=−2/(1+x²).
At x=±1, the derivative does not exist.
Answer
y′ = 2/(1+x²) for |x|<1; y′ = −2/(1+x²) for |x|>1. It is not differentiable at x=±1.
Question 10
Find dy/dx if y = tan⁻¹((3x−x³)/(1−3x²)), where −1/√3 < x < 1/√3.
Steps only
Let θ=tan⁻¹x. The fraction is tan 3θ.
The interval gives −π/2<3θ<π/2, so y=3θ.
y′=3/(1+x²).
Answer
3/(1+x²).
Question 11
Find dy/dx if y = cos⁻¹((1−x²)/(1+x²)), where 0 < x < 1.
Steps only
Let θ=tan⁻¹x. The fraction is cos 2θ.
Here 0<2θ<π/2, so y=2θ.
y′=2/(1+x²).
Answer
2/(1+x²).
Question 12
Find dy/dx if y = sin⁻¹((1−x²)/(1+x²)), where 0 < x < 1.
Steps only
Let θ=tan⁻¹x. The fraction is cos 2θ = sin(π/2−2θ).
The angle lies in the principal range of sin⁻¹, so y=π/2−2θ.
y′=−2/(1+x²).
Answer
−2/(1+x²).
Question 13
Find dy/dx if y = cos⁻¹(2x/(1+x²)), where −1 < x < 1.
Steps only
Let θ=tan⁻¹x. Then 2x/(1+x²)=sin 2θ=cos(π/2−2θ).
The angle is in [0,π], so y=π/2−2θ.
y′=−2/(1+x²).
Answer
−2/(1+x²).
Question 14
Find dy/dx if y = sin⁻¹(2x√(1−x²)), where −1/√2 < x < 1/√2.
Steps only
Let θ=sin⁻¹x. Then 2x√(1−x²)=sin 2θ.
The interval ensures 2θ is in the principal range of sin⁻¹, so y=2θ.
y′=2/√(1−x²).
Answer
2/√(1−x²).
Question 15
Find dy/dx if y = sec⁻¹(1/(2x²−1)), where 0 < x < 1/√2.
Steps only
Let θ=cos⁻¹x. Then 1/(2x²−1)=sec 2θ.
The interval gives π/2<2θ<π, in the principal range of sec⁻¹ for this input; y=2θ.
y′=−2/√(1−x²).
Answer
−2/√(1−x²).
Exercise 5.4
Textbook pages 130
Question 1
Differentiate y = eˣ/sin x.
Steps only
y′=(eˣsin x−eˣcos x)/sin²x.
Answer
eˣ(sin x − cos x)/sin²x, where x ≠ nπ for n∈ℤ.
Question 2
Differentiate y = e^(sin⁻¹x).
Steps only
y′=e^(sin⁻¹x)·1/√(1−x²).
Answer
e^(sin⁻¹x)/√(1−x²), for −1<x<1.
Question 3
Differentiate y = e^(x³).
Steps only
y′=e^(x³)·3x².
Answer
3x²e^(x³).
Question 4
Differentiate y = sin(tan⁻¹(e^(−x))).
Steps only
Let u=e^(−x). Then y′=cos(tan⁻¹u)·[−u/(1+u²)].
Answer
−e^(−x)/(1+e^(−2x))^(3/2).
Question 5
Differentiate y = log(cos(eˣ)).
Steps only
y′=[−sin(eˣ)·eˣ]/cos(eˣ).
Answer
−eˣ tan(eˣ), wherever cos(eˣ)>0.
Question 6
Differentiate y = eˣ + e^(x²) + e^(x³) + e^(x⁴) + e^(x⁵).
Steps only
Differentiate term by term using the chain rule.
Answer
eˣ + 2xe^(x²) + 3x²e^(x³) + 4x³e^(x⁴) + 5x⁴e^(x⁵).
Question 7
Differentiate y = √(e^(√x)), where x>0.
Steps only
Rewrite y=e^(√x/2).
y′=e^(√x/2)/(4√x).
Answer
e^(√x/2)/(4√x), for x>0.
Question 8
Differentiate y = log(log x), where x>1.
Steps only
y′=[1/log x]·[1/x].
Answer
1/(x log x), for x>1.
Question 9
Differentiate y = cos x/log x, where x>0.
Steps only
y′=[−sin x·log x−cos x·(1/x)]/(log x)².
Answer
−(x sin x log x + cos x)/(x(log x)²), for x>0 and x≠1.
Question 10
Differentiate y = cos(log x + eˣ), where x>0.
Steps only
y′=−sin(log x+eˣ)·(1/x+eˣ).
Answer
−(1/x + eˣ) sin(log x + eˣ), for x>0.
Exercise 5.5 · Logarithmic differentiation
Textbook pages 178–179
Question 1
Differentiate y = cos x · cos 2x · cos 3x.
Steps only
y′/y=−tan x−2 tan 2x−3 tan 3x.
Answer
y′=−cos x cos 2x cos 3x (tan x+2 tan 2x+3 tan 3x).
Question 2
Differentiate y = (x−1)²(x−2)²/[(x−3)³(x−4)⁴(x−5)⁵].
Steps only
y′/y=2/(x−1)+2/(x−2)−3/(x−3)−4/(x−4)−5/(x−5).
Answer
y′=y[2/(x−1)+2/(x−2)−3/(x−3)−4/(x−4)−5/(x−5)], where y is the given function.
Question 3
Differentiate y = (log x)^(cos x).
Steps only
y′/y=cos x/(x log x)−sin x·log(log x).
Answer
y′=(log x)^(cos x)[cos x/(x log x)−sin x·log(log x)], for x>1.
Question 4
Differentiate y = x^(x−2 sin x).
Steps only
y′/y=(1−2cos x)log x + 1−2sin x/x.
Answer
y′=x^(x−2sin x)[(1−2cos x)log x+1−2sin x/x], for x>0.
Question 5
Differentiate y = (x+3)²(x+4)³(x+5)⁴.
Steps only
y′/y=2/(x+3)+3/(x+4)+4/(x+5).
Answer
y′=(x+3)(x+4)²(x+5)³(9x²+70x+133).
Question 6
Differentiate y = (x+1/x)^x + x^(1+1/x).
Steps only
d/dx[(x+1/x)^x]=(x+1/x)^x[log(x+1/x)+(x²−1)/(x²+1)].
d/dx[x^(1+1/x)]=x^(1+1/x)[1/x+(1−log x)/x²].
Answer
y′=(x+1/x)^x[log(x+1/x)+(x²−1)/(x²+1)]+x^(1+1/x)[1/x+(1−log x)/x²], for x>0.
Question 7
Differentiate y = (log x)^x + x^(log x).
Steps only
d/dx[(log x)^x]=(log x)^x[log(log x)+1/log x].
d/dx[x^(log x)]=x^(log x)·2log x/x.
Answer
y′=(log x)^x[log(log x)+1/log x]+2x^(log x)log x/x, for x>1.
Question 8
Differentiate y = (sin x)^x + (sin x)^(cos x).
Steps only
d/dx[(sin x)^x]=(sin x)^x[log(sin x)+x cot x].
d/dx[(sin x)^(cos x)]=(sin x)^(cos x)[−sin x·log(sin x)+cos x cot x].
Answer
y′=(sin x)^x[log(sin x)+x cot x]+(sin x)^(cos x)[−sin x·log(sin x)+cos x cot x], where sin x>0.
Question 9
Differentiate y = x^(sin x) + (sin x)^(cos x).
Steps only
d/dx[x^(sin x)]=x^(sin x)[cos x·log x+sin x/x].
d/dx[(sin x)^(cos x)]=(sin x)^(cos x)[−sin x·log(sin x)+cos x cot x].
Answer
y′=x^(sin x)[cos x·log x+sin x/x]+(sin x)^(cos x)[−sin x·log(sin x)+cos x cot x], for x>0 and sin x>0.
Question 10
Differentiate y = x^(x cos x) + (x²+1)/(x²−1).
Steps only
d/dx[x^(x cos x)]=x^(x cos x)[cos x(1+log x)−x sin x log x].
d/dx[(x²+1)/(x²−1)]=−4x/(x²−1)².
Answer
y′=x^(x cos x)[cos x(1+log x)−x sin x log x]−4x/(x²−1)², for x>0 and x≠1.
Question 11
Differentiate y = (x cos x)^x + (x sin x)^(1/x).
Steps only
d/dx[(x cos x)^x]=(x cos x)^x[log(x cos x)+1−x tan x].
d/dx[(x sin x)^(1/x)]=(x sin x)^(1/x)[x cot x+1−log(x sin x)]/x².
Answer
y′=(x cos x)^x[log(x cos x)+1−x tan x]+(x sin x)^(1/x)[x cot x+1−log(x sin x)]/x², wherever x cos x>0 and x sin x>0.
Question 12
If x^y + y^x = 1, find dy/dx.
Steps only
x^y[y/x+(dy/dx)log x]+y^x[(x/y)(dy/dx)+log y]=0.
Answer
dy/dx=−[y x^(y−1)+y^x log y]/[x^y log x+x y^(x−1)], for x>0 and y>0 where the denominator is nonzero.
Question 13
If y^x = x^y, find dy/dx.
Steps only
x log y=y log x.
y′(x/y−log x)=y/x−log y.
Answer
dy/dx = y(y−x log y)/[x(x−y log x)], for x>0 and y>0 where the denominator is nonzero.
Question 14
If (cos x)^y = (cos y)^x, find dy/dx.
Steps only
y log(cos x)=x log(cos y).
y′[log(cos x)+x tan y]=log(cos y)+y tan x.
Answer
dy/dx=[log(cos y)+y tan x]/[log(cos x)+x tan y], wherever the logarithms are defined and the denominator is nonzero.
Question 15
If xy = e^(x−y), find dy/dx.
Steps only
log x+log y=x−y.
1/x+y′/y=1−y′.
Answer
dy/dx=y(x−1)/[x(y+1)], for x>0 and y>0.
Question 16
For f(x)=(1+x)(1+x²)(1+x⁴)(1+x⁸), find f′(x) and f′(1).
Steps only
(1+x)(1+x²)(1+x⁴)(1+x⁸)=1+x+x²+…+x¹⁵.
f′(x)=1+2x+3x²+…+15x¹⁴; f′(1)=120.
Answer
f′(x)=1+2x+3x²+…+15x¹⁴, and f′(1)=120.
Question 17
Differentiate (x²−5x+8)(x³+7x+9) using the product rule, expansion, and logarithmic differentiation. Do the methods agree?
Steps only
Product rule: (2x−5)(x³+7x+9)+(x²−5x+8)(3x²+7).
Expanded product: x⁵−5x⁴+15x³−26x²+11x+72.
Answer
All three methods give 5x⁴−20x³+45x²−52x+11.
Question 18
For differentiable functions u, v and w, show that d(uvw)/dx = u′vw+uv′w+uvw′ in two ways: repeated product rule and logarithmic differentiation.
Steps only
Repeated product rule: (uvw)′=((uv)w)′=(u′v+uv′)w+uv w′.
Logarithmic method: (uvw)′=uvw(u′/u+v′/v+w′/w).
Answer
(uvw)′=u′vw+uv′w+uvw′.
Exercise 5.6 · Parametric differentiation
Textbook pages 137
Question 1
If x=2at² and y=at⁴, find dy/dx without eliminating t.
Steps only
dx/dt=4at and dy/dt=4at³.
dy/dx=(4at³)/(4at)=t².
Answer
dy/dx=t².
Question 2
If x=a cos θ and y=b cos θ, find dy/dx.
Steps only
dx/dθ=−a sin θ; dy/dθ=−b sin θ.
dy/dx=(−b sin θ)/(−a sin θ)=b/a.
Answer
dy/dx=b/a (where the parametric ratio is defined).
Question 3
If x=sin t and y=cos 2t, find dy/dx.
Steps only
dx/dt=cos t; dy/dt=−2sin 2t.
dy/dx=−2sin 2t/cos t=−4sin t, where cos t≠0.
Answer
dy/dx=−4 sin t.
Question 4
If x=4t and y=4/t, find dy/dx.
Steps only
dx/dt=4; dy/dt=−4/t².
dy/dx=(−4/t²)/4=−1/t².
Answer
dy/dx=−1/t², for t≠0.
Question 5
If x=cos θ−cos 2θ and y=sin θ−sin 2θ, find dy/dx.
Steps only
dx/dθ=−sin θ+2sin 2θ.
dy/dθ=cos θ−2cos 2θ.
dy/dx=(cos θ−2cos 2θ)/(2sin 2θ−sin θ).
Answer
dy/dx=(cos θ−2cos 2θ)/(2sin 2θ−sin θ), where the denominator is nonzero.
Question 6
If x=a(θ−sin θ) and y=a(1+cos θ), find dy/dx.
Steps only
dx/dθ=a(1−cos θ); dy/dθ=−a sin θ.
dy/dx=−sin θ/(1−cos θ)=−cot(θ/2).
Answer
dy/dx=−cot(θ/2), where 1−cos θ≠0.
Question 7
If x=sin³t/√(cos 2t) and y=cos³t/√(cos 2t), find dy/dx.
Steps only
Log-differentiate the coordinate functions: (dx/dt)/x=3cot t+tan 2t and (dy/dt)/y=−3tan t+tan 2t.
Divide the rates and simplify with triple-angle identities: dy/dx=−cot 3t.
Answer
dy/dx=−cot 3t, where the coordinates and dx/dt are defined and dx/dt≠0.
Question 8
If x=a(cos t+log tan(t/2)) and y=a sin t, find dy/dx.
Steps only
dx/dt=a(−sin t+cosec t)=a cos²t/sin t.
dy/dt=a cos t.
dy/dx=(a cos t)/(a cos²t/sin t)=tan t.
Answer
dy/dx=tan t, where the expressions are defined and dx/dt≠0.
Question 9
If x=a sec θ and y=b tan θ, find dy/dx.
Steps only
dx/dθ=a sec θ tan θ; dy/dθ=b sec²θ.
dy/dx=(b/a)(sec θ/tan θ)=(b/a)cosec θ.
Answer
dy/dx=(b/a)cosec θ, where sec θ and the ratio are defined.
Question 10
If x=a(cos θ+θ sin θ) and y=a(sin θ−θ cos θ), find dy/dx.
Steps only
dx/dθ=aθ cos θ; dy/dθ=aθ sin θ.
dy/dx=tan θ.
Answer
dy/dx=tan θ, where θ cos θ≠0.
Question 11
If x=√(a^(sin⁻¹t)) and y=√(a^(cos⁻¹t)), show that dy/dx=−y/x.
Steps only
Write x=a^((sin⁻¹t)/2) and y=a^((cos⁻¹t)/2).
dx/dt=x log(a)/(2√(1−t²)); dy/dt=−y log(a)/(2√(1−t²)).
(dy/dt)/(dx/dt)=−y/x.
Answer
dy/dx=−y/x, for a>0, a≠1, −1<t<1.
Exercise 5.7 · Second-order derivatives
Textbook pages 139–140
Question 1
Find the second derivative of y=x²+3x+2.
Steps only
y′=2x+3.
y″=2.
Answer
y″=2.
Question 2
Find the second derivative of y=x²⁰.
Steps only
y′=20x¹⁹.
y″=20·19x¹⁸=380x¹⁸.
Answer
y″=380x¹⁸.
Question 3
Find the second derivative of y=x cos x.
Steps only
y′=cos x−x sin x.
y″=−sin x−sin x−x cos x=−x cos x−2sin x.
Answer
y″=−x cos x−2 sin x.
Question 4
Find the second derivative of y=log x.
Steps only
y′=1/x.
y″=−1/x².
Answer
y″=−1/x², for x>0.
Question 5
Find the second derivative of y=x³ log x.
Steps only
y′=3x²log x+x².
y″=6xlog x+3x+2x=x(6log x+5).
Answer
y″=x(5+6 log x), for x>0.
Question 6
Find the second derivative of y=eˣ sin 5x.
Steps only
y′=eˣ(sin 5x+5cos 5x).
y″=eˣ(10cos 5x−24sin 5x).
Answer
y″=2eˣ(5 cos 5x−12 sin 5x).
Question 7
Find the second derivative of y=e⁶ˣ cos 3x.
Steps only
y′=e⁶ˣ(6cos 3x−3sin 3x).
y″=e⁶ˣ(27cos 3x−36sin 3x).
Answer
y″=9e⁶ˣ(3 cos 3x−4 sin 3x).
Question 8
Find the second derivative of y=tan⁻¹x.
Steps only
y′=1/(1+x²).
y″=−2x/(1+x²)².
Answer
y″=−2x/(1+x²)².
Question 9
Find the second derivative of y=log(log x).
Steps only
y′=1/(x log x).
y″=−(log x+1)/(x²(log x)²).
Answer
y″=−(1+log x)/[x²(log x)²], for x>1.
Question 10
Find the second derivative of y=sin(log x).
Steps only
y′=cos(log x)/x.
y″=−sin(log x)/x²−cos(log x)/x².
Answer
y″=−[sin(log x)+cos(log x)]/x², for x>0.
Question 11
If y=5 cos x−3 sin x, prove that y″+y=0.
Steps only
y′=−5sin x−3cos x.
y″=−5cos x+3sin x=−y.
Therefore y″+y=0.
Answer
y″+y=0.
Question 12
If y=cos⁻¹x, find d²y/dx² in terms of y alone.
Steps only
dy/dx=−1/√(1−x²)=−cosec y, because x=cos y and 0<y<π.
d²y/dx²=−(d/dy cosec y)(dy/dx)=−cot y·cosec²y.
Answer
d²y/dx²=−cot y·cosec²y.
Question 13
If y=3 cos(log x)+4 sin(log x), show that x²y″+xy′+y=0.
Steps only
Let u=log x. Then xy′=dy/du and x²y″+xy′=d²y/du².
d²y/du²=−3cos u−4sin u=−y.
Thus x²y″+xy′+y=0.
Answer
x²y″+xy′+y=0, for x>0.
Question 14
If y=Aeᵐˣ+Beⁿˣ, show that y″−(m+n)y′+mn y=0.
Steps only
y′=mAeᵐˣ+nBeⁿˣ.
y″=m²Aeᵐˣ+n²Beⁿˣ.
For each exponential, the coefficient is m²−(m+n)m+mn=0 or n²−(m+n)n+mn=0.
Answer
y″−(m+n)y′+mn y=0.
Question 15
If y=500e⁷ˣ+600e⁻⁷ˣ, show that y″=49y.
Steps only
y′=3500e⁷ˣ−4200e⁻⁷ˣ.
y″=24500e⁷ˣ+29400e⁻⁷ˣ=49(500e⁷ˣ+600e⁻⁷ˣ)=49y.
Answer
y″=49y.
Question 16
If eʸ(x+1)=1, show that d²y/dx²=(dy/dx)².
Steps only
Take logarithms: y+log(x+1)=0.
y′=−1/(x+1); y″=1/(x+1)².
Therefore y″=(y′)².
Answer
y″=(y′)², for x>−1.
Question 17
If y=(tan⁻¹x)², show that (x²+1)²y″+2x(x²+1)y′=2.
Steps only
y′=2 tan⁻¹x/(1+x²).
y″=2/(1+x²)²−4x tan⁻¹x/(1+x²)².
Substitution gives 2−4x tan⁻¹x+4x tan⁻¹x=2.
Answer
(x²+1)²y″+2x(x²+1)y′=2.
Miscellaneous Exercise on Chapter 5
Textbook pages 144–146
Question 1
Differentiate y=(3x²−9x+5)⁹.
Steps only
y′=9(3x²−9x+5)⁸(6x−9).
y′=27(3x²−9x+5)⁸(2x−3).
Answer
y′=27(3x²−9x+5)⁸(2x−3).
Question 2
Differentiate y=sin³x+cos⁶x.
Steps only
y′=3sin²x cos x−6cos⁵x sin x.
Factor 3sin x cos x.
Answer
y′=3sin x cos x(sin x−2cos⁴x).
Question 3
Differentiate y=(5x)^(3cos 2x), for x>0.
Steps only
log y=3cos 2x·log(5x).
y′/y=−6sin 2x·log(5x)+3cos 2x/x.
Answer
y′=(5x)^(3cos 2x)[3cos 2x/x−6sin 2x·log(5x)].
Question 4
Differentiate y=sin⁻¹(x√x), for 0≤x≤1.
Steps only
x√x=x^(3/2).
y′=[1/√(1−x³)]·(3/2)x^(1/2).
Answer
y′=3√x/[2√(1−x³)], for 0≤x<1 (right derivative at x=0); it is not finite at x=1.
Question 5
Differentiate y=cos⁻¹(x/2)/√(2x+7), for −2<x<2.
Steps only
u=cos⁻¹(x/2), v=(2x+7)^(−1/2).
u′=−1/√(4−x²), v′=−(2x+7)^(−3/2).
y′=u′v+uv′.
Answer
y′=−1/[√(4−x²)√(2x+7)]−cos⁻¹(x/2)/(2x+7)^(3/2).
Question 6
Differentiate y=cot⁻¹[(√(1+sin x)+√(1−sin x))/(√(1+sin x)−√(1−sin x))], for 0<x<π/2.
Steps only
√(1+sin x)=cos(x/2)+sin(x/2).
√(1−sin x)=cos(x/2)−sin(x/2).
The ratio is cot(x/2), so y=x/2 and y′=1/2.
Answer
y′=1/2.
Question 7
Differentiate y=(log x)^(log x), for x>1.
Steps only
log y=(log x)·log(log x).
y′/y=[1+log(log x)]/x.
Answer
y′=(log x)^(log x)[1+log(log x)]/x.
Question 8
Differentiate y=cos(a cos x+b sin x), where a and b are constants.
Steps only
u=a cos x+b sin x, so u′=−a sin x+b cos x.
y′=−sin(u)u′.
Answer
y′=(a sin x−b cos x)sin(a cos x+b sin x).
Question 9
Differentiate y=(sin x−cos x)^(sin x−cos x), for π/4<x<3π/4.
Steps only
Let u=sin x−cos x>0.
log y=u log u, so y′/y=u′[1+log u] and u′=cos x+sin x.
Answer
y′=(sin x−cos x)^(sin x−cos x)(cos x+sin x)[1+log(sin x−cos x)].
Question 10
Differentiate y=xˣ+xᵃ+aˣ+aᵃ, where a>0 is constant and x>0.
Steps only
(xˣ)′=xˣ(1+log x).
(xᵃ)′=a x^(a−1), (aˣ)′=aˣ log a, and (aᵃ)′=0.
Answer
y′=xˣ(1+log x)+a x^(a−1)+aˣ log a.
Question 11
Differentiate y=x^(x²−3)+(x−3)^(x²), for x>3.
Steps only
For u=x^(x²−3), u′/u=2x log x+(x²−3)/x.
For v=(x−3)^(x²), v′/v=2x log(x−3)+x²/(x−3).
y′=u′+v′.
Answer
y′=x^(x²−3)[2x log x+(x²−3)/x]+(x−3)^(x²)[2x log(x−3)+x²/(x−3)].
Question 12
If y=12(1−cos t) and x=10(t−sin t), find dy/dx for −π/2<t<π/2.
Steps only
dy/dt=12sin t.
dx/dt=10(1−cos t).
dy/dx=(6/5)sin t/(1−cos t)=(6/5)cot(t/2).
Answer
dy/dx=(6/5)cot(t/2), for t≠0. At t=0 the parametric ratio is undefined.
Question 13
Find dy/dx if y=sin⁻¹x+sin⁻¹√(1−x²), for 0<x<1.
Steps only
For 0<x<1, sin⁻¹√(1−x²)=cos⁻¹x.
y=sin⁻¹x+cos⁻¹x=π/2, so y′=0.
Answer
dy/dx=0.
Question 14
If x√(1+y)+y√(1+x)=0, where −1<x<1, prove that dy/dx=−1/(1+x)².
Steps only
Let p=√(1+x), q=√(1+y). The equation becomes (p+q)(pq−1)=0.
Since p+q>0, pq=1, so (1+x)(1+y)=1 and y=−x/(1+x).
Differentiate to get dy/dx=−1/(1+x)².
Answer
dy/dx=−1/(1+x)².
Question 15
If (x−a)²+(y−b)²=c², where c>0, show that [(1+(dy/dx)²)^(3/2)]/(d²y/dx²) is constant on each graph branch and independent of a and b.
Steps only
Differentiate: y′=−(x−a)/(y−b).
Then y″=−c²/(y−b)³ and 1+(y′)²=c²/(y−b)².
The ratio is −c·sgn(y−b), constant on each branch.
Answer
The ratio equals −c on the upper semicircle (y>b) and +c on the lower semicircle (y<b); each value is constant and independent of a and b.
Question 16
If cos y=x cos(a+y), where cos a≠±1, prove that dy/dx=cos²(a+y)/sin a.
Steps only
Differentiate: −sin y·y′=cos(a+y)−x sin(a+y)y′.
Use x=cos y/cos(a+y), so sin y−x sin(a+y)=−sin a/cos(a+y).
Therefore y′=cos²(a+y)/sin a.
Answer
dy/dx=cos²(a+y)/sin a.
Question 17
If x=a(cos t+t sin t) and y=a(sin t−t cos t), find d²y/dx².
Steps only
dx/dt=a t cos t; dy/dt=a t sin t.
dy/dx=tan t.
d²y/dx²=(d/dt tan t)/(dx/dt)=1/(a t cos³t).
Answer
d²y/dx²=1/(a t cos³t), where a t cos t≠0.
Question 18
If f(x)=|x|³, show that f″(x) exists for every real x and find it.
Steps only
For x>0, f=x³ so f″=6x.
For x<0, f=−x³ so f″=−6x.
At x=0, f′(0)=0 and f″(0)=limₕ→0[f′(h)−f′(0)]/h=0.
Answer
f″(x)=6|x| for all real x.
Question 19
Starting from sin(A+B)=sin A cos B+cos A sin B and using differentiation, obtain the cosine addition formula.
Steps only
Differentiate the sine addition identity with respect to A.
cos(A+B)=cos A cos B−sin A sin B.
Answer
cos(A+B)=cos A cos B−sin A sin B.
Question 20
Does there exist a function continuous everywhere but not differentiable at exactly two points? Justify your answer.
Steps only
Absolute-value functions are continuous, so their sum is continuous.
The slopes are −2 for x<0, 0 for 0<x<1, and 2 for x>1.
The slope changes at 0 and 1 only.
Answer
Yes. One example is f(x)=|x|+|x−1|; it is continuous everywhere and is not differentiable exactly at x=0 and x=1.
Question 21
Let y be the determinant with rows [f(x),g(x),h(x)], [l,m,n], and [a,b,c], where l,m,n,a,b,c are constants. Prove that y′ is the determinant obtained by replacing the first row by [f′(x),g′(x),h′(x)].
Steps only
Expand along the first row: y=f(x)(mc−nb)−g(x)(lc−na)+h(x)(ln−ma).
Differentiate; the cofactors are constants.
Collect as the determinant with the differentiated first row.
Answer
y′=| f′(x) g′(x) h′(x); l m n; a b c |.
Question 22
If y=e^(a cos⁻¹x), for −1≤x≤1, show that (1−x²)y″−xy′−a²y=0.
Steps only
y′=−a y/√(1−x²).
y″=a²y/(1−x²)−a x y/(1−x²)^(3/2).
Substitution gives a²y−a x y/√(1−x²)+a x y/√(1−x²)−a²y=0.
Answer
(1−x²)y″−xy′−a²y=0 for −1<x<1, where the derivatives are defined.