Exercise 1

Differentiate the following trigonometric functions:

f(x) = sin \frac{1}{2}x

f(x) = cos (7 - 2x)

f(x) = 3 tan 2x

f(x) = sec(5x + 2)

f(x) = cos \frac{x + 1}{x - 1}

 

The best Maths tutors available
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
1st lesson free!
Myriam
5
5 (15 reviews)
Myriam
£20
/h
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
1st lesson free!
Myriam
5
5 (15 reviews)
Myriam
£20
/h
First Lesson Free>

Exercise 2

Differentiate the following inverse trigonometric functions:

f(x) = arcsin (1 - 2x^2)

f(x) = arcsin \sqrt{x^2 - 4}

f(x) = arccos e^x

f(x) = arctan \sqrt{x}

f(x) = arctan \frac{1 + x}{1 - x}

f(x) = arctan \frac{x}{\sqrt{1 - x^2}}

 

Exercise 3

Differentiate the following functions using the chain rule:

f(x) = ln(sin x)

f(x) = ln (cos2x)

f(x) = ln tan(1 - x)

f(x) = ln \sqrt{\frac{1 + sinx}{1 - sinx}}

f(x) = sin\sqrt{ln (1 - 3x)}

f(x) = tan (sin \sqrt{5x})

f(x) = sin^2 (cos 2x)

f(x) = cos (cos (cos x))

Exercise 4

Differentiate the following functions:

f(x) = (sin x) ^ {cos x}

f(x) = \sqrt[x^2] {arccos x}

Exercise 5

Find the successive derivatives:

f(x) = 3x^4 + 5x^2 + 2x - 5

f(x) = lnx

f(x) = sin x

f(x) = e^{-3x}

Exercise 6

Differentiate the following functions:

x^2y - xy^2 + y^2 = 7

x^2 sin(x + y) - 5ye^{x} = 3

 

Solution of exercise 1

Differentiate the following trigonometric functions:

1  f(x) = sin \frac{1}{2}x

f'(x) = \frac{1}{2} cos \frac{1}{2}x

 

2  f(x) = cos(7 - 2x)

f'(x) = - (-2) \cdot sin(7 - 2x) = 2 \cdot sin (7 - 2x)

 

3  f(x) = 3 tan 2x
f'(x) = 6(1 + tan^2 2x)
4 f(x) = sec(5x + 2)
f'(x) = 5 tan (5x + 2) \cdot sec(5x + 2)
 

5  f(x) = cos\frac{x + 1}{x - 1}

f'(x) = - \frac{x - 1- (x + 1)}{(x - 1)^2} sin \frac{x + 1}{x - 1} = \frac{2}{(x - 1)^2} \cdot sin \frac{x + 1}{x - 1}

 

Solution of exercise 2

Differentiate the following inverse trigonometric functions:

f(x) = arcsin ( 1 - 2x^2)

f'(x) = \frac{-4x}{\sqrt{1 - (1 - 2x^2)^2}}

 

f(x) = arcsin \sqrt{x^2 - 4}

f'(x) = \frac{1}{\sqrt{(x^2 - 4)}} \cdot \frac{2x}{2\sqrt{x^2 - 4}} = \frac{x}{\sqrt{5 - x^2} \cdot \sqrt {x^2 - 4}}

 

3  f(x) = arccos e^x

f'(x) = - \frac{e^x}{\sqrt{1 - e^{2x}}}

 

4  f(x) = arctan \sqrt{x}
f'(x) = \frac{1}{1 + x} \frac{1}{2 \sqrt{x}} = \frac{1}{2\sqrt{x} (1 + x)}
5  f(x) = arctan \frac{1 + x}{1 - x}
f'(x) = \frac{1}{1 + (\frac{1 + x}{1 - x})^2} \cdot \frac {1 - x + 1 + x}{(1 - x)^2}

= \frac{1}{1 + \frac{(1 + x)^2} {(1 - x)}^2} \cdot \frac{2}{(1 - x)^2} = \frac{2}{(1 - x)^2 + (1 + x)^2}

= \frac{2}{1 - 2x + x^2 + 1 + 2x + x^2} = \frac{2}{2 + 2x^2}

= \frac{1}{1 + x^2}

 

f(x) = arctan \frac{x}{\sqrt{1 - x^2}}

f'(x) = \frac{1}{1 + (\frac{x}{\sqrt{1 - x^2})^2}}  \cdot \frac{\sqrt{1 - x^2} -x \cdot \frac{-2x}{2 \sqrt{1 - x^2}}} { (\sqrt{1 - x^2})^2}

= \frac{1}{\frac{1 - x^2 + x^2}{1 - x^2}} \cdot \frac {2 (1 - x^2) + 2x^2}{\frac {2 \sqrt{1 - x^2}} { 1 - x^2}}

= \frac{1}{\frac{1}{1 - x^2}} \cdot \frac{2}{\frac{2 \sqrt{1 - x^2}} { 1 - x^2}}

= (1 - x^2) \cdot \frac{1}{(1 - x^2) \sqrt{1 - x^2}}

= \frac{1}{\sqrt{1 - x^2}}

 

Solution of exercise 3

Differentiate the following functions using the chain rule:

1  f(x) = ln(sin x)

f'(x) = \frac{cosx}{sinx} = cot x

 

f(x) = ln (cos 2x)

f'(x) = \frac{-2sin 2x}{cos 2x} = -2 tan 2x

 

3 f(x) = ln tan (1 - x)

f'(x) = \frac{1 + tan ^2 (1 - x)}{tan (1 - x)}

 

4  f(x) = ln \sqrt{\frac{1 + sin x}{1 - sin x}}

f(x) = \frac{1}{2}[ln (1 + sin x) - ln (1 - sin x)]

f'(x) = \frac{1}{2} (\frac{cos x}{1 + sin x} - \frac{-cos x} { 1 - sinx})

= \frac{1}{2} \cdot (\frac{cos x}{1 + sin x} - \frac{-cos x} { 1 - sin x})

= \frac{1}{2} \cdot \frac{cos x - sin x cos x + cos x + sin x cos x} { 1 - sin^2 x}

\frac{1}{2} \cdot \frac{2 cos x}{cos^2 x} = \frac{1}{cos x} = secx

 

5  f(x) = sin \sqrt{ln (1 - 3x)}

f'(x) = cos \sqrt{ln (1 - 3x)} \cdot \frac{1}{2 \sqrt{ln (1 - 3x)}} \cdot \frac{1}{1 - 3x} \cdot (-3)

 

6  f(x) = tan (sin \sqrt{5x})
f'(x) = [ 1 + tan^2 (sin \sqrt{5x})] \cdot cos \sqrt{5x} \cdot \frac{1}{2 \sqrt{5x}} \cdot 5
7  f(x) = cos ( cos (cos x))

f'(x) = -sin (cos (cos x)) \cdot sin (cos x) \cdot sin x

 

 

Solution of exercise 4

Differentiate the following functions:

1  f(x) = (sin x)^ {cos x}

y = (sin x)^ {cos x}

lny = ln(sin x) ^ {cos x}             ln y = cos x ln (sin x)

\frac{y'}{y} = -sin x ln (sin x) + cos x \frac{cos x}{sin x}

f'(x) = (- sin x ln (sin x) + \frac{cos^2 x}{sin x}) (sin x) ^ {cos x}

 

f(x) = \sqrt[x^2] {arccos x}

y = (arc cos x) ^{\frac{1}{x^2}}

ln y = \frac{1}{x^2} ln arc cos x

\frac{y'}{y} = - \frac{2}{x^3} ln arc cos x - \frac{1}{x^2} \frac{1}{arc cos x} \frac{1}{\sqrt{1 - x^2}}

f(x) = - \frac{1}{x^2} \sqrt[x^2]{arc cos x} ( \frac{2}{x} ln arc cos x + \frac{1}{\sqrt{1 - x^2} arc cos x})

 

Solution of exercise 5

Find the successive derivatives:

f(x) = 3x^4 + 5x^2 + 2x - 5

f'(x) = 12x^3 + 10x + 2

f ''(x) = 36x^2 + 10

f '''(x) = 72x

f^{IV} (x) = 72

f^{V} (x) = 0

 

 

2  f(x) = ln x

f '(x) = \frac{1}{x}

f ''(x) = \frac{-1}{x^2}

f '''(x) = \frac{2}{x^3}

f ^ {IV} (x) = \frac{2 \cdot 3}{x^4}

f^n(x) = (-1) ^ {n - 1} \frac{(n - 1)!} {x^n}

 

 

3  f(x) = sin x

f '(x) = cos x = sin (\frac{\pi}{2} + x)

f ''(x) = -sin x = - [-sin (\pi + x)] = sin (2 \cdot \frac{\pi}{2} + x)

f '''(x) = -cos x = -sin (\frac{\pi}{2} + x) = - [-sin (3 \cdot \frac{\pi}{2} + x)] = sin (3 \cdot \frac{\pi}{2} + x)

f^n(x) = sin (\frac{n \cdot \pi}{2} + x )

 

 

4  f(x) = e ^ {-3x}

f '(x) = -3 e^ {-3x}

f ''(x) = 9 \cdot e ^ {-3x}

f ''' (x) = -27 e ^{-3x}

f^n(x) = (-3)^n \cdot e ^ {-3x}

 

Solution of exercise 6

Differentiate the following functions:

x^2y - xy ^2 + y^2 = 7

2xy + x^2y' - (y^2 + 2xyy') + 2yy' = 0

2xy +x^2y' - y^2 - 2xyy' + 2yy' = 0

x^2y' - 2xyy' + 2yy' = -2xy + y^2

y'(x^2 - 2xy + 2y) = y^2 - 2xy

y' = \frac{y^2 - 2xy}{x^2 - 2xy + 2y}

 

 

2   x^2 sin (x + y) - 5ye^x = 3

y' = \frac{-[2x sin (x + y) + x^2 cos(x + y) - 5y e^x]}{x^2 cos (x + y) - 5e^x}

y' = \frac{2x sin (x + y) + x^2 cos (x + y) - 5ye^x}{-x^2 cos (x + y) + 5e^x}

 

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 5.00/5 - 1 vote(s)
Loading...

Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.