Chapter 1 MATHEMATICAL PHYSICS 1.1 TRIGONOMETRY Radian: It is the angle subtended by an arc of a circle of length equal to radius of circle at the centre of circle 1 rad = 57.296 0 If ‘ l ’ represents length of an arc of circle of radius ‘R’ subtending an angle ′ 𝜃 ′ at the centre. Then 𝑙 = 𝑅𝜃 ( 𝜃 is in Radians) Angle subtended by a complete circle at centre is 2 𝜋 𝑐 or 360 0 2 𝜋 𝑐 = 36 0 0 1 𝑐 = 36 0 0 2 𝜋 or 1 0 = 2 𝜋 𝑐 360 𝑥 0 = 𝑥 × 2 𝜋 𝑐 360 and 𝑥 𝑐 = 𝑥 × 36 0 0 2 𝜋 Problems: 1. 6 0 0 = _ _ _ 𝑐 2. 12 0 0 = _ _ _ 𝑐 3. 4 5 0 = _ _ _ 𝑐 4. 30 0 = _ _ _ 𝑐 5. 5 𝜋 𝑐 6 = _ _ _ 0 6. 2 𝜋 𝑐 3 = _ _ _ 0 7. 𝜋 𝑐 2 = _ _ _ 0 Trigonometric Ratio: Let Opposite side to ′ 𝜃 ′ is 𝐵𝐶 , adjacent side to ′ 𝜃 ′ is 𝐴𝐵 and hypotenuse is 𝐴𝐶 • sin θ = opp Hyp = BC AC = x √ x 2 + y 2 • cos θ = adj Hyp = AB AC = y √ x 2 + y 2 • tan θ = sin θ cos θ = opp adj = BC AB = x y • cot θ = 1 tan θ = cos θ sin θ = adj opp = AB BC = y x • sec θ = 1 cos θ = Hyp adj = AC AB = √ x 2 + y 2 y • cosec θ = 1 sin θ = Hyp opp = AC BC = √ x 2 + y 2 x Trigonometric Identities • sin 2 θ + cos 2 θ = 1 sin θ = √ 1 − cos 2 θ cos θ = √ 1 − sin 2 θ • sec 2 θ − tan 2 θ = 1 • cos e c 2 θ − cot 2 θ = 1 T rigonometric Ratio of standard angle: Ratio/θ 0 0 30 0 45 0 60 0 90 0 sin θ 0 1 2 1 √ 2 √ 3 2 1 cos θ 1 √ 3 2 1 √ 2 1 2 0 tan θ 0 1 √ 3 1 √ 3 N. D cosec θ N. D 2 √ 2 2 √ 3 1 sec θ 1 2 √ 3 √ 2 2 N. D cot θ N. D √ 3 1 1 √ 3 0 Range of T rigonometric Ratios • sin θ ∈ [ − 1 , 1 ] • cos θ ∈ [ − 1 , 1 ] • tan θ ∈ ( − ∞ , ∞ ) • cot θ ∈ ( − ∞ , ∞ ) • sec θ ∈ ( − ∞ , − 1 ] ∪ [ 1 , ∞ ) • cos e c θ ∈ ( − ∞ , − 1 ] ∪ [ 1 , ∞ ) ASTC rule of trigonometry • If angle lies in first quadrant a ll trigonometric ratios are positive. • If angle lies in second quadrant, the sin 𝜃 and cosec 𝜃 are positive • If angle lies in third quadrant, the tan 𝜃 and cot 𝜃 are positive • If angle lies in fourth quadrant, the cos 𝜃 and sec 𝜃 are positive Allied angles : • If angles are in the form of 90 ± 𝜃 , 270 ± 𝜃 sin ⇌ cos tan ⇌ cot sec ⇌ cos e c • Angles in the form of 180 ± 𝜃 , 360 ± 𝜃 sin → sin cot → cot cos → cos sec → sec tan → tan cosec → cosec Example: 𝑠𝑖𝑛 ( 180 + 𝜃 ) = − 𝑠𝑖𝑛 𝜃 Since 180 + 𝜃 lies in 3 rd quadrant and sine is negative in 3 rd quadrant. 90 – θ 90 + θ 180 – θ 180 + θ 270 – θ 270+ θ 360 - θ 360+ θ sin cos tan cot sec cosec Trigonometric Ratio of Negative Angles • sin ( − θ ) = − sin θ • cos ( − θ ) = + cos θ • sec ( − θ ) = sec θ • cos e c ( − θ ) = − cos e c θ • tan ( − θ ) = − tan θ • cot ( − θ ) = − cot θ Some basic formulae • sin ( x + y ) = sin x cos y + cos x sin y • sin ( x − y ) = sin x cos y − cos x sin y • cos ( x + y ) = cos x cos y − sin x sin y • cos ( x − y ) = cos x cos y + sin x sin y • sin 2 x = 2 sin x cos x • cos 2 x = cos 2 x − sin 2 x = 2 cos 2 x − 1 = 1 − 2 sin 2 x • cos 2 x = 1 + cos2x 2 • sin 2 x = 1 − cos 2 x 2 • sin x + sin y = 2 sin ( x + y 2 ) cos ( x − y 2 ) • sin x − sin y = 2 cos ( x + y 2 ) sin ( x − y 2 ) • cos x + cos y = 2 cos ( x + y 2 ) cos ( x − y 2 ) • cos x − cos y = − 2 sin ( x + y 2 ) sin ( x − y 2 ) Solutions of equation General solution of • sin θ = 0 if θ = n π , n ∈ Z • sin θ = 1 if θ = 2n π + π 2 , n ∈ Z • sin θ = - 1 if θ = 2n π − π 2 , n ∈ Z • cos θ = 0 if θ = ( 2n + 1 ) π 2 , n ∈ Z • cos θ= - 1 if θ = ( 2n + 1 ) π , n ∈ Z • cos θ = +1 if θ = 2n π , n ∈ Z Graphs of trigonometric functions : (i) sin θ (ii) cos θ Maximum & Minimum value of a cos θ + b sin θ a cos θ + b sin θ = √ a 2 + b 2 ( a √ a 2 + b 2 cos θ + b √ a 2 + b 2 sin θ ) ( Since sin ∝ = a √ a 2 + b 2 and cos ∝ = b √ a 2 + b 2 ) a cos θ + b sin θ = √ a 2 + b 2 ( sin ∝ cos θ + cos ∝ sin θ ) a cos θ + b sin θ = √ a 2 + b 2 ( ( sin ∝ + θ ) ) , where tan ∝ = a b Since maximum value of 𝑠𝑖𝑛 ( ∝ + 𝜃 ) is +1, maximum value of 𝑎 𝑐𝑜𝑠 𝜃 + 𝑏 𝑠𝑖𝑛 𝜃 = + √ 𝑎 2 + 𝑏 2 Since minimum value of 𝑠𝑖𝑛 ( ∝ + 𝜃 ) is - 1, m in imum value of 𝑎 𝑐𝑜𝑠 𝜃 + 𝑏 𝑠𝑖𝑛 𝜃 = − √ 𝑎 2 + 𝑏 2 Problems: 1. Convert the following angles into radian A) 30 0 B) 45 0 C) 60 0 D) 90 0 E) 120 0 F) 135 0 G) 150 0 H) 180 0 I) 270 0 J) 360 0 2. Covert the following angles into degrees A) 𝜋 𝑐 10 B) 𝜋 𝑐 9 C) 4 𝜋 𝑐 3 D) 𝜋 𝑐 4 E) 3 𝜋 𝑐 2 F) 11 𝜋 𝑐 6 G) 5 𝜋 𝑐 3 3. If 0 0 < 𝜃 < 9 0 0 , 𝑠𝑖𝑛 𝜃 = 2 3 . Find the values of other trigonometric ratios for ‘ 𝜃 ’? 4. If 18 0 0 < 𝜃 < 27 0 0 , 𝑠𝑖𝑛 𝜃 = − 1 3 . Find the values of other trigonometric ratios for ‘ 𝜃 ’? 5. Find the values of following A) 𝑠𝑖𝑛 ( 18 0 0 + 4 5 0 ) B) 𝑠𝑖𝑛 ( 27 0 0 + 3 0 0 ) C) 𝑡𝑎𝑛 ( 9 0 0 + 6 0 0 ) D) 𝑐𝑜𝑠 𝑒 𝑐 ( 9 0 0 − 3 0 0 ) E) 𝑐𝑜𝑠 ( 36 0 0 − 4 5 0 ) F) 𝑐𝑜𝑡 ( 18 0 0 + 3 0 0 ) G) 𝑡𝑎𝑛 ( − 4 5 0 ) H) 𝑐𝑜𝑠 𝑒 𝑐 ( 27 0 0 + 4 5 0 ) I) 𝑠𝑒𝑐 ( 9 0 0 + 3 0 0 ) J) 𝑐𝑜𝑠 ( − 13 5 0 ) 6. What is the maximum and minimum value of 2 𝑐𝑜𝑠 𝜃 + 3 𝑠𝑖𝑛 𝜃 ? ANSWER KEY 1. A) 𝜋 6 B) 𝜋 4 C) 𝜋 3 D) 𝜋 2 E) 2 𝜋 3 F) 3 𝜋 4 G) 5 𝜋 6 H) 𝜋 H) 3 𝜋 2 J) 2 𝜋 2. A) 18 0 B) 20 0 C) 240 0 D) 45 0 E) 270 0 F) 330 0 G) 300 0 3. cos θ = √ 5 3 ; sec θ = 3 √ 5 ; θ = 3 2 ; tan θ = 2 √ 5 ; cot θ = √ 5 2 4. cos θ = − 2 √ 2 3 ; sec θ = − 3 2 √ 2 ; tan θ = 1 2 √ 2 ; cot θ = 2 √ 2 ; cosec θ = − 3 5. A) − 1 √ 2 B) − √ 3 2 C) − 1 √ 3 D) 2 √ 3 E) 1 √ 2 F) √ 3 G) - 1 𝐻 ) − √ 2 I) - 2 J) − 1 √ 2 6. Maxim um value is √ 2 2 + 3 2 ; Minimum value is - √ 2 2 + 3 2 1.2 COORDINATE GEOMETRY • (x, y) represents a point which is at x - units from y - axis and y - units away from x - axis. • A point on x - axis will be of the form (x, 0) • A point on y - axis will be in the form of (0, y) • (0, 0) represents origin which is point of intersection of x & y axes. • ( 𝑥 , 𝑦 ) → ( + , + ) → lies i n first q uadrat • ( 𝑥 , 𝑦 ) → ( − , + ) → lies in second q uadrant • ( 𝑥 , 𝑦 ) → ( − , − ) → lies in third q uadrant • ( 𝑥 , 𝑦 ) → ( + , − ) → lies in fourth q uadrant Linear Equation in two variables: An equation of the form ax + by + c = 0 represents a straight line on a graph. Slope of line: Slope of a line represents inclination of a line w ith respect to positive x - axis. If ‘θ’ represents angle made by line ax + by +c = 0 with x - axis in anti - clockwise direction, the slope of the line is given by m = tan θ 𝑚 = − coefficient of x coefficient of y = − 𝑎 𝑏 ; Here “ m ” is slope of straight - line ax + by +c =0 • m is positive if 0 0 < θ < 90 0 • m is negative if 90 0 < θ < 180 0 • m is ‘0’ if θ = 0 0 • m is not defined if θ = 90 0 • Slope of a line parallel to x – axis is ‘zero’ ( ∵ θ = 0 0 ) • Slope of a line parallel to y - axis is not defined ( ∵ θ = 90 0 ) • Slope of a line passing through points (x 1 , y 1 ) and (x 2 , y 2 ) is m = y 2 − y 1 x 2 − x 1 Intercepts of a straight line X – intercept: X - intercept of a line is the x - co - ordinate of the point of intersection of the line with x - axis. Y - intercept: Y - intercept of line is the y - coordinate of the point of intersection of the line with y - axis. Equations of different form: • Equation of x - axis is y = 0 • Equation of y - axis is x = 0 • Equation of the form x = c passes through the points with x - coordinate as ‘c’ • Equation of the form y = c passes through the points with y - coordinate as ‘c’ • An equation of the form y = mx is a straight line passing through origin with slope ‘m’. • Equation of line having slope ‘m’ and passing ( 𝑥 1 , 𝑦 1 ) is ( 𝑦 − 𝑦 1 ) = 𝑚 ( 𝑥 − 𝑥 1 ) • Equation of line passing through the points ( 𝑥 1 , 𝑦 1 ) and ( 𝑥 2 , 𝑦 2 ) is ( 𝑦 − 𝑦 1 ) = ( 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 ) ( 𝑥 − 𝑥 1 ) 𝑜𝑟 ( 𝑦 − 𝑦 2 ) = ( 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 ) ( 𝑥 − 𝑥 2 ) • Equation of a line having slope ‘m’ and y - intercept ‘c’ is y = mx + c. • Equation of a line with ‘a’ and ‘b’ as x and y - intercepts respectively is 𝑥 𝑎 + 𝑦 𝑏 = 1 • Distance between two points (x 1 , y 1 , z 1 ) and ( 𝑥 2 , 𝑦 2 , 𝑧 2 ) = √ ( 𝑥 2 − 𝑥 1 ) 2 + ( 𝑦 2 − 𝑦 1 ) 2 + ( 𝑧 2 − 𝑧 1 ) 2 Some common graphs: Equation Graph y = mx y = mx + c ( 𝑚 , 𝑐 ) → ( + , + ) ( 𝑚 , 𝑐 ) → ( + , − ) ( 𝑚 , 𝑐 ) → ( − , + ) ( 𝑚 , 𝑐 ) → ( − , − ) 𝒚 = 𝒄 𝒙 𝟐 𝒐𝒓 𝒙 𝟐 = 𝟒𝒂𝒚 𝒙 = 𝒄 𝒚 𝟐 𝒐𝒓 𝒚 𝟐 = 𝟒𝒂𝒙 xy = c; where c is positive Problems 1. Represent the following points on x - y co - ordinate system. A) (0, 3) B) ( - 4, 0) C) (0, - 2) D) (2, 3) E) ( - 1, 2) F) ( - 3, - 2) G) (2, - 4) H) (3, 0) 2. What is the sign of slope of lines ′ 𝐿 1 ′&′ 𝐿 2 ′ mentioned in graph? 3. What is the slope of a line passing through points (1, 4) and ( - 2, 3)? 4. What is the slope of a line parallel to x – axis? 5. What is the slope of straight line 2 𝑥 + 3 𝑦 = 5 . Draw the rough graph of the above linear equation. 6. Draw the following graphs. A) 𝑥 = 3 B) 𝑥 = − 2 C) 𝑦 = 4 D) 𝑦 = − 1 E) 𝑦 = 𝑥 F) 𝑦 = − 𝑥 G) 𝑦 = 3 𝑥 I) 𝑦 = 2 𝑥 + 3 I) 𝑦 = − 𝑥 − 4 J) 𝑦 = 𝑥 2 K) 𝑥 = 𝑦 2 7. What is the distance between points? A) (2, 3) and (4, 5) B) (0, 0) and (1, 1) C) (4, 5) and ( - 2, - 3) 8. Which of the following lines passes through origin? A) 𝑦 = 4 𝑥 + 3 B) 𝑦 = 2 𝑥 C) 𝑦 = 3 𝑥 − 5 9. Find the co - ordinates of point of intersection of 2 𝑥 + 3 𝑦 − 5 = 0 and 4 𝑥 − 2 𝑦 + 3 = 0 ANSWER KEYS 1. 2. Slope of ‘L 1 ’ is ‘positive’; Slope of ‘L 2 ’ is ‘Negative’ 3. Slope of line = 1 3 4. Zero 5. Slope = − 2 3 6. Graph 7. A) 2 √ 2 B) √ 2 C) 10 8. Y = 2x 9. ( 1 16 , 13 8 ) 1.3 GEOMETRY Equilateral Triangle: • All angles are equal to 60 0 • All sides are equal. • Area = √ 3 4 𝑎 2 , 𝑎 → length of each side. • Length of median = √ 3 2 ( 𝑎 ) • A median and corresponding a l titude coincide. • A median is also an angular Bisector. • In all triangles, centroid divides median in the ratio of 2:1 AG: GD = 2:1 where ‘AD’ is the median and ‘G’ is centroid. • Sum of all angles in a triangle is equal to 180 0 • In a triangle sum of any two sides must be greater than 3 rd side. Difference of any two sides must be smaller than third side. • Area of a triangle = 1 2 × 𝑏 × ℎ , Where ‘b’ is base length and ‘h’ is height. Isosceles Tr iang le: • Two sides are equal in length and their correspon ding angles are equal. • ABC is an isosceles triangle with A B = AC. Here BC is called Base. • Median drawn to base is an a l titude to that side. Square: • All sides are equal in length. • All angles are equal to 90 0 • Length of a diagonal = √ 2 a. • Diagonals are equal in length. • Diagonals of a square are perpendicular bisectors of each other. • Diagonals are also angular bisectors. Rectangle: • All angles are equal to 90 0 • Opposite pairs of sides are equal in length and parallel. • Length of diagonal = √ 𝑎 2 + 𝑏 2 Parallelogram: • Both pairs of opposite sides are parallel. • Opposite angles are equal • Sum of adjacent angles = 180 0 • Area = base × height. Circle: • Diameter = 2(Radius) • Area = 𝜋 𝑅 2 = 𝜋 𝐷 2 4 Where, D is diameter. • Perimeter (or) circumference of circle = 2 𝜋𝑅 = 𝜋𝐷 • Length of an arc, 𝑙 = 𝑅𝜃 = angle subtended by arc at centre in radians. Volume of some geometrical objects: • Volume of a cube = a 3 • Volume of a Cuboid = l bh • Volume of a cylinder = 𝜋 𝑅 2 ℎ • Volume of a Sphere = 4 3 𝜋 𝑅 3 • Volume of a cone = 1 3 𝜋 𝑅 2 ℎ Surface area of some geometrical objects: • Surface are a of Cube = 6a 2 • Surface are a of Cuboid = 2 ( l b + bh + l h) • Surface are a of Cylinder = 2 𝜋𝑟 ℎ + 2 𝜋 𝑅 2 • Surface are a of Sphere = 4 𝜋 𝑅 2 • Surface area of Cone = 𝜋𝑟𝑙 + 𝜋 𝑅 2 NOTE : Distance of vertex of a regular Hexagon from centre of hexagon is equal to side length of hexagon. Parallel Lines: • (1, 3), (2, 4), (6, 8), (5, 7) are opposite angles. Opposite angles are equal. • (2, 6), (1, 5); (3, 7); (4, 8) are four pairs of corresponding angles. Corresponding angles are equal. • ∠1 = ∠3 ; ∠2 = ∠4 ; ∠6 = ∠8 ; ∠5 = ∠7 • ∠2 = ∠6 ; ∠1 = ∠5 ; ∠3 = ∠7 ; ∠4 = ∠8 • ∠1 = ∠7 ; ∠4 = ∠6 ; ∠3 = ∠5 ; ∠2 = ∠8 • ∠1 + ∠2 = ∠3 + ∠4 = ∠6 + ∠5 = ∠7 + ∠8 = 18 0 0 Congruency of Triangles: Two triangles are said to be congruent if corresponding sides and angles are equal. Two triangles can be congruent in following ways: • SSS congruency • SAS congruency • ASA congruency • RHS Congruency Similarity of Triangles: Two t riangles are said to be similar if the pairs of angles are equal In the given two angles ∠ 𝐴 = ∠ 𝐷 ; ∠ 𝐵 = ∠ 𝐸 ; ∠ 𝐶 = ∠ 𝐹 ∴ 𝛥𝐴𝐵𝐶 ≈ 𝛥𝐷𝐸𝐹 In two similar triangles, ratio of corresponding sides will be equal. 𝐴𝐵 𝐷𝐸 = 𝐵𝐶 𝐸𝐹 = 𝐶𝐴 𝐹𝐷 Problems 1. In 𝛥𝐴𝐵𝐶 , 𝐴𝐵 = 𝐴𝐶 , 𝐴𝐷 ⊥ 𝐵𝐶 . Find the value of ′ 𝛼 ′ 2. If side length of an equilateral triangle is ‘a’. Find the distance of one of the vertices from centroid of the triangle. 3. ABCD is a square of side length ‘a’. What is the length of BD and BE, BF? Where E and F are midpoints of ‘AD’ and ‘DC’ respectively. ANSWER KEYS 1. 𝜶 = 𝟔 𝟎 𝟎 2. 𝒂 √ 𝟑 3. 𝑩𝑫 = √ 𝟐 𝒂 , 𝑩𝑬 = 𝑩𝑭 = √ 𝟓 𝒂 𝟐 1.4 ALGEBRA Solution of linear and quadratic equations: • Solution of two linear equations 𝑎 1 𝑥 + 𝑏 1 𝑦 + 𝑐 1 = 0 and 𝑎 2 𝑥 + 𝑏 2 𝑦 + 𝑐 2 = 0 is ( 𝑥 , 𝑦 ) = ( 𝑏 1 𝑐 2 − 𝑏 2 𝑐 1 𝑎 1 𝑏 2 − 𝑎 2 𝑏 1 , 𝑐 1 𝑎 2 − 𝑐 2 𝑎 1 𝑎 1 𝑏 2 − 𝑎 2 𝑏 1 ) • Roots of a Quadratic equation 𝑎 𝑥 2 + 𝑏𝑥 + 𝑐 = 0 are 𝑥 = − 𝑏 ± √ 𝑏 2 − 4 𝑎𝑐 2 𝑎 Sum of roots = − 𝑏 𝑎 ; P roduct of roots = 𝑐 𝑎 Sum of first ‘n’ natural number • ∑ 𝑛 = 1 + 2 + 3 + + 𝑛 = 𝑛 ( 𝑛 + 1 ) 2 • ∑ 𝑛 2 = 1 2 + 2 2 + 3 2 + + 𝑛 2 = 𝑛 ( 𝑛 + 1 ) ( 2 𝑛 + 1 ) 6 • ∑ 𝑛 3 = 1 3 + 2 3 + 3 3 + + 𝑛 3 = ( 𝑛 ( 𝑛 + 1 ) 2 ) 2 Arithmetic and geometric sequence : • ( 𝑎 ) + ( 𝑎 + 𝑑 ) + ( 𝑎 + 2 𝑑 ) + 𝑛𝑡𝑒𝑟𝑚𝑠 = 𝑛 2 ( 2 𝑎 + ( 𝑛 − 1 ) 𝑑 ) • 𝑎 + 𝑎 𝑟 + 𝑎 𝑟 2 + + 𝑎 𝑟 𝑛 − 1 = 𝑎 ( 𝑟 𝑛 − 1 ) 𝑟 − 1 • If r < < 1, then 𝑎 + 𝑎𝑟 + 𝑎 𝑟 2 + infinite terms= 𝑎 1 − 𝑟 • ( 1 + 𝑥 ) 𝑛 ≈ 1 + 𝑛𝑥 ; when 𝑥 < < 1 Algebraic identities: • ( 𝑎 + 𝑏 ) 2 = 𝑎 2 + 𝑏 2 + 2 𝑎𝑏 • ( 𝑎 − 𝑏 ) 2 = 𝑎 2 + 𝑏 2 − 2 𝑎𝑏 • ( 𝑎 + 𝑏 ) 2 − ( 𝑎 − 𝑏 ) 2 = 4 𝑎𝑏 • ( 𝑎 + 𝑏 ) 3 = 𝑎 3 + 𝑏 3 + 3 𝑎𝑏 ( 𝑎 + 𝑏 ) • ( 𝑎 − 𝑏 ) 3 = 𝑎 3 − 𝑏 3 − 3 𝑎𝑏 ( 𝑎 − 𝑏 ) Problems: 1. Find the point of intersection of y = 3 and 4x – 3y =2. 2. Find the roots of quadratic equation 𝑥 2 − 5 𝑥 + 6 = 0 . Also find sum and product of roots. 3. Calculate the sum of first 10 natural numbers. 4. Calculate the sum of squares of first 20 natural numbers. 5. Calculate the sum of cubes of first 5 natural numbers. 6. 1 + 1 2 + 1 4 + 1 8 + 1 16 + + ∞ 𝑡𝑒𝑟𝑚𝑠 = _ _ _ _ _ _ _ _ _ _ _ _ _ _ 7. If 𝑎 + 𝑏 = 9 and 𝑎 𝑏 = 77 4 . Calculate the value of ( 𝑎 − 𝑏 ) 2 1.4 ALGEBRA 1. ( 11 4 , 3 ) 2. Roots are 2 (or) 3 Sum of roots = 5Product of roots = 6 3. 55 4. 2870 5. 225 6. 2 7. 4 1.5 DIFFERENTIATION Basic Formula • 𝑑 𝑑𝑥 (constant) = 0 • 𝑑 𝑑𝑥 ( 𝑥 𝑛 ) = 𝑛 𝑥 𝑛 − 1 • 𝑑 𝑑𝑥 ( 𝑐 𝑢 ) = 𝑐 𝑑 𝑑𝑥 ( 𝑢 ) where , ‘c’ is a constant. • 𝑑 𝑑𝑥 ( 𝑒 𝑥 ) = 𝑒 𝑥 • 𝑑 𝑑𝑥 ( 𝑙𝑛𝑥 ) = 1 𝑥 • 𝑑 𝑑𝑥 ( 𝑠𝑖𝑛 𝑥 ) = 𝑐𝑜𝑠 𝑥 • 𝑑 𝑑𝑥 ( 𝑐𝑜𝑠 𝑥 ) = − 𝑠𝑖𝑛 𝑥 • 𝑑 𝑑𝑥 ( 𝑡𝑎𝑛 𝑥 ) = 𝑠𝑒𝑐 2 𝑥 • 𝑑 𝑑𝑥 ( 𝑠𝑒𝑐 𝑥 ) = 𝑠𝑒𝑐 𝑥 𝑡𝑎𝑛 𝑥 • 𝑑 𝑑𝑥 ( 𝑐𝑜𝑡 𝑥 ) = − 𝑐𝑜𝑠 𝑒 𝑐 2 𝑥 • 𝑑 𝑑𝑥 ( 𝑐𝑜𝑠 𝑒 𝑐𝑥 ) = − 𝑐𝑜𝑠 𝑒 𝑐𝑥 𝑐𝑜𝑡 𝑥 • Graphically 𝑑𝑦 𝑑𝑥 gives slope of tangent drawn at any point on the graph of function y = f (x) • If ‘u’ and ‘v’ are function of ‘x’ ❖ 𝑑 𝑑𝑥 ( 𝑢 + 𝑣 ) = 𝑑 𝑑𝑥 ( 𝑢 ) + 𝑑 𝑑𝑥 ( 𝑣 ) ❖ 𝑑 𝑑𝑥 ( 𝑢 − 𝑣 ) = 𝑑 𝑑𝑥 ( 𝑢 ) − 𝑑 𝑑𝑥 ( 𝑣 ) ❖ 𝑑 𝑑𝑥 ( 𝑢 𝑣 ) = 𝑢 𝑑 𝑑𝑥 ( 𝑣 ) + 𝑣 𝑑 𝑑𝑥 ( 𝑢 ) ❖ 𝑑 𝑑𝑥 ( 𝑢 𝑣 ) = 𝑣 𝑑 𝑑𝑥 ( 𝑢 ) − 𝑢 𝑑 𝑑𝑥 ( 𝑣 ) 𝑣 2 Chain Rule: • 𝑑 𝑑𝑥 ( 𝑓 ( 𝑦 ) ) = 𝑑 𝑑𝑦 ( 𝑓 ( 𝑦 ) ) × 𝑑𝑦 𝑑𝑥 where ‘y’ is a function of ‘x’. Maximum and Minimum value of a function: • A function f (x) has maximum value at 𝑥 = 𝑥 1 if 𝑓 ′ ( 𝑥 1 ) = 0 & 𝑓 ′′ ( 𝑥 1 ) < 0 , then maximum value of f(x) is 𝑓 ( 𝑥 1 ) • A function f(x) has minimum value at 𝑥 = 𝑥 1 , if 𝑓 ′ ( 𝑥 1 ) = 0 𝑓 ′′ ( 𝑥 1 ) > 0 , t hen minimum value of f(x) is 𝑓 ( 𝑥 1 ) • 𝑑𝑦 𝑑𝑥 also represents rate of change of ‘y’ w ith respect to ‘x’ Problems: 1. Differentiate the following functions w ith respect to ‘x’? A) 𝑥 2 B) 𝑥 5 3 C) 2 𝑐𝑜𝑠 𝑥 D) 1 𝑥 4 E) e 3 F) 2 𝑠𝑖𝑛 𝑥 + 3 𝑐𝑜𝑠 𝑥

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