Survival-Guide-2 - Tutorial guide for Differential Calculus and Integral Calculus PDF

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Tutorial guide for Differential Calculus and Integral Calculus...

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MESSAGE TO THE USER: The aim of this guide is to serve as assistance in learning and is not to undermine the efforts of the teachers. The contents of this guide are not error-free so we ask that if errors are found please inform us right away to continuously improve our guide. - CE Council Intellectual Committee (2018-2019)

CIVIL ENGINEERING GUIDE

SOLID MENSURATION

PLANE FIGURES

REGULAR POLYGON

Given: a, b, c

- A regular polygon of n sides can be divided into m congruent isosceles triangles. Area 𝐴 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) Perimeter 𝑃 =𝑎+𝑏+𝑐

𝑠=

RECTANGLE

where:

𝑛𝑠 2

180° 𝑛

4𝑡𝑎𝑛(

)

SQUARE

where: s=length of 1 side n=number of sides

TRIANGLE -A polygon that has 3 sides and 3 vertices Given: b and h

Area 1 𝐴 = 2 𝑏ℎ Perimeter 𝑃 =𝑎+𝑏+𝑐

Where: B = base H = height

Given: a, b, ϴ

Area 1 𝐴 = 𝑎𝑏 sin 𝜃 2 Perimeter 𝑃 =𝑎+𝑏+𝑐

𝑎+𝑏+𝑐 2

-Is a parallelogram in which the interior angles are all right angles. Diagonal 𝑑 = √𝑏 2 + ℎ2 Perimeter 𝑃 = 2𝑏 + 2ℎ Area 𝐴 = 𝑏ℎ

a = apothem 𝜃 = central angle s = length of side Central Angle 360° 𝜃= ; n = no. of sides 𝑛 Apothem 360° 𝑠 𝜃 = tan = tan 2 𝑛 2𝑎 Solving for a, 𝑠 𝑎= 180° 2 tan ( 𝑛 ) Area 1 where: P=perimeter 𝐴 = 2 𝑃𝑎 a=apothem 𝐴=

Where:

Where: 𝜃 = angle between given sides

-A special type of rectangle in which all sides are equal. Diagonal 𝑑 = 𝑎√2 Perimeter 𝑃 = 4𝑎 Area 𝐴 = 𝑎2

RHOMBUS

-Is a parallelogram in which all sides are equal. Where: h = height b = length of one side

Diagonal

𝑑1 2 = 2𝑏 2 (1 − cos 𝜃) 𝑑2 2 = 2𝑏 2 (1 + cos 𝜃)

𝑃 = 4𝑏 1 If given d1 & d2, 𝐴 = 𝑑1 𝑑2 2 If given b & h, 𝐴 = 𝑏ℎ If given b & ϴ, 𝐴 = 𝑏 2 sin 𝜃 Perimeter

TRAPEZOID -Is a quadrilateral with one pair of parallel sides

Area 1 𝐴 = (𝑏 + 𝑎 )(ℎ) 2 Perimeter 𝑃 =𝑎+𝑏+𝑐+𝑑

CIRCLE

-A circle is a set of points, each of which is equidistant from a fixed point call a center

Area

𝐴 = 𝜋𝑟 2 𝑜𝑟 Perimeter 𝑃 = 2𝜋𝑟 Arc length

𝑠 = 𝑟𝜃

𝜋𝐷 4

Where: 𝜃 = central angle s = arc length

Area 1 𝐴 = 𝑟𝑠 2 Perimeter 𝑃 = 𝑟(2 + 𝜃)

𝑠 = 𝑟𝜃

𝑟=

𝑟=

𝑟=

𝑛𝑠 2

180°

4 tan( 𝑛 )

𝑎𝑏𝑐 4𝑎

𝑎𝑏𝑐

4√𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)

where: 𝑠 =

𝑎+𝑏+𝑐 2

where: 𝑠 =

𝑎+𝑏+𝑐 2

𝐴 𝑠

𝑟=√

Where: L = chord Segment Area 1 𝐴 = 𝑟 2 (𝜃 − sin 𝜃) 2 Perimeter 𝑃 = 𝑠+𝐿 ;

where: s = length of one side r = radius 𝜃 = central angle (rad)

“Inscribed regular polygon” 𝑠 = 𝑟√𝑥(1 − cos 𝜃) 𝑃 = 𝑛𝑠 𝐴=

2

Sector

Circumscribed & Inscribed Polygons -A polygon is inscribed in a circle if the vertices of the polygon lie on the circle -A polygon is circumscribed about the circle if each side of the polygon is tangent to the circle. Examples:

(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐) 𝑠

ELLIPSE -is a conic with an eccentricity that is less than 1 𝐿 = 𝑟 √2(1 − cos 𝜃) 𝐴 = 𝜋𝑎𝑏

𝑃 = 2𝜋√−

where: a = major radius b = minor

𝑎 2 +𝑏 1 2

PARABOLA -is a conic section in which the eccentricity is equal to 1

Area of Parabola = 3 𝑏ℎ 2

Area of Spandrel = 𝑏ℎ 3 𝑃=

1

√𝑏2 +16ℎ2 2

+ 8ℎ ln [ 𝑏2

4ℎ+√𝑏2 +16ℎ2 𝑏

]

PRISMS

Solid FiguresREGULAR PYRAMID

-They can be described as a three dimensional solid formed by dragging a polygon through space to a certain height without rotating or tilting the polygon.

-is a right pyramid whose base is a regular polygon

Where: B = length of one side E = lateral edge Where: h = height e = lateral edge b = length of one side l = slant height B = base area Lateral Surface Area: 𝐿𝑆𝐴 = 𝑃𝑒 Total Surface Area: B is the area of one base 𝑇𝑆𝐴 = 𝐿𝑆𝐴 + 2𝐵 Volume: B is the area of the base; h is the height 𝑉 = 𝐵ℎ

CYCLINDER

-is a solid bounded by a closed cylindrical surface and two parallel planes cutting all the elements of the surface.

Lateral Surface Area: P = perimeter 1 𝐿𝑆𝐴 = 𝑃𝐿 2 Total Surface Area: B = area of the base 𝑇𝑆𝐴 = 𝐵 = 𝐿𝑆𝐴 Volume: 1 𝑉 = 𝑏ℎ 3

CONE

-is the solid formed by one nappe of a closed surface and a plane cutting all the elements of the line Where: l = slant height h = height r = radius

Lateral Surface Area: 𝐿𝑆𝐴 = 2𝜋𝑟ℎ Total Surface Area: 𝑇𝑆𝐴 = 2𝜋(𝑟 + ℎ) Volume: 𝑉 = 𝐵ℎ = 𝜋𝑟 2 ℎ

Lateral Surface Area 𝐿𝑆𝐴 = 𝜋𝑟𝑙 Total Surface Area 𝑇𝑆𝐴 = 𝜋𝑟(𝑟 + 𝑙 ) Volume 1 𝑉 = (𝜋𝑟 2 ℎ) 3

FRUSTUM OF A RIGHT CIRCULAR CONE

PRISMATOID

-is a portion of the cone enclosed by its base, a section that is parallel to the base and the conical surface included between the base of the cone and the parallel section.

-is a polyhedron where vertices all lie in two parallel planes.

Where: l = slant height h = height r1= radius of the upper base r2= radius of the lower base

𝑙 = √ℎ2 + (𝑟2 + 𝑟1 )2 Later Surface Area: C is the circumference 𝐿𝑆𝐴 = 𝜋 (𝑟1 + 𝑟2 )𝑙 Total Surface Area 𝑇𝑆𝐴 = 𝜋(𝑟1 + 𝑟2 )𝑙 + 𝜋𝑟12 + 𝜋𝑟22 Volume: 𝜋ℎ(𝑟12 + 𝑟22 + 𝑟1 𝑟2 ) 𝑉= 3

FRUSTUM OF A REGULAR PYRAMID

-it is a polyhedron enclosed by the lateral surface, the base of the pyramid and the section of the pyramid

Volume: B1, B2, M are all areas 1 𝑣 = ℎ(𝐵1 + 𝐵2 + 4𝑀) 6

-

SPHERE -a sphere is a three-dimensional solid bounded by a surface consisting of all points equidistant from an interior point called the center. Where: O = center R = Radius Surface Area 𝑆 = 4𝜋𝑅 2 Volume 4 𝑉 = 𝜋𝑟 2 3

INSCRIBED SOLIDS

-is a solid of maximum volume placed inside another solid of fixed volume, with their edges of surfaces touching each other. “A cylinder inscribed in a sphere”

Where: e = lateral edge l = slant height h = height Lateral Surface Area: P is the perimeter 1 𝐿𝑆𝐴 = (𝑃1 + 𝑃2 )𝑙 2 Total Surface Area: B1 & B2 are the areas 𝑆𝐴 = 𝐿𝑆𝐴 + 𝐵1 + 𝐵2 Volume: ℎ(𝐵1 + 𝐵2 √𝐵1 𝐵2 ) 𝑣= 3

Analysis: D = diameter of sphere 2r = diameter of cylinder h = height of cylinder 𝐷 = √(2𝑟)2 + ℎ2

“A sphere inscribed in a cube”

Analysis: D = is the diameter of the sphere S = length of an edge of the cube

PAPPUS-GULDINUS THEOREMS Theorem 1: The surface of the revolution is defined as the surface generated by rotating a plane curve through 360° about an axis 𝑠 = 2𝜋 ∑ 𝐿𝑑 Where: L = length of an element d = centroidal distance from axis of revolution Example

Theorem 2: The solid of revolution is the solid formed by rotating a plane through 360° about an axis 𝑉 = 2𝜋 ∑ 𝐴𝑑

Where: D = centroidal distance form axis of revolution A = area of the plane Example

DIFFERENTIAL CALCULUS

LIMITS AND CONTINUITY •

DEFINITION -Let f be a function defined at every number in some open interval containing a, except possibly at the number a itself. The limit of f(x) as x approaches a is L, written as: lim 𝑓(𝑥) = 𝐿 if the following statement is true: 𝑥→𝑎

Given any ∈> 0, however small, there exists a 𝛿 > 0 such that if 0 < |𝑥 − 𝑎| < 𝛿 then |𝑓(𝑥) − 𝐿| 0 we know that xr stays in the denominator. As we increase x, xr will also increase. So we will have a constant divided by an increasingly large number and so the result will also be increasingly small. b. If r is a positive rational number, c is any real number and xr is defined for x 0 and if 𝑓(𝑥) → 0 through positive values of 𝑓(𝑥), 𝑔(𝑥) = +∞ lim 𝑥→𝑎 𝑓(𝑥) If 𝑐 > 0 and if 𝑓(𝑥) → 0 through negative values of 𝑓(𝑥), 𝑔(𝑥) = −∞ lim 𝑥→𝑎 𝑓(𝑥)

If 𝑐 < 0 and if 𝑓(𝑥) → 0 through positive values of 𝑓(𝑥), 𝑔(𝑥) = −∞ lim 𝑥→𝑎 𝑓(𝑥) (iv) If 𝑐 < 0 and if 𝑓(𝑥) → 0 through negative values of 𝑓(𝑥), 𝑔(𝑥) lim = +∞ 𝑥→𝑎 𝑓(𝑥) NOTE: The theorem will remain valid if 𝑥 → 𝑎 is replaced by 𝑥 → 𝑎 + 𝑜𝑟 𝑥 → 𝑎 − c. SUMS INVOLVING INFINITE LIMITS (i) If lim 𝑓(𝑥) = +∞, and lim 𝑔(𝑥) = 𝑐, (iii)

𝑥→𝑎

𝑥→𝑎

where c is any constant, then

lim [𝑓(𝑥) + 𝑔(𝑥)] = +∞ (ii) If𝑥→𝑎 lim 𝑓(𝑥) = −∞, and lim 𝑔(𝑥) = 𝑐, where c is any constant, then 𝑥→𝑎

𝑥→𝑎

lim[𝑓( 𝑥) + 𝑔(𝑥)] = −∞

𝑥→𝑎

NOTE: The theorem will remain valid if 𝑥 → 𝑎 is replaced by 𝑥 → 𝑎 + 𝑜𝑟 𝑥 → 𝑎 −

d. PRODUCTS INVOLVING INFINITE LIMITS If lim 𝑓(𝑥) = +∞, and lim 𝑔(𝑥) = 𝑐, 𝑥→𝑎

𝑥→𝑎

where c is any constant except 0, then (i) If 𝑐 > 0, lim[𝑓(𝑥)●𝑔(𝑥)] = +∞ (ii)

𝑥→𝑎

If 𝑐 < 0, lim[𝑓(𝑥)●𝑔(𝑥)] = −∞ 𝑥→𝑎

NOTE: The theorem will remain valid if 𝑥 → 𝑎 is replaced by 𝑥 → 𝑎 + 𝑜𝑟 𝑥 → 𝑎 − e. If lim 𝑓(𝑥) = −∞, and lim 𝑔(𝑥) = 𝑐, 𝑥→𝑎

𝑥→𝑎

where c is any constant except 0, then (i) If 𝑐 > 0, lim [𝑓(𝑥)●𝑔(𝑥)] = −∞

(ii)

f.

𝑥→𝑎

NOTE: The theorem will remain valid if 𝑥 → 𝑎 is replaced by 𝑥 → 𝑎 + 𝑜𝑟 𝑥 → 𝑎 − DEFINITION OF A VERTICAL ASYMPTOTE The line 𝑥 = 𝑎 is a vertical asymptote of the function f if at least one of the following statements is true. (i) (ii) (iii) (iv)

•

𝑥→𝑎

If 𝑐 < 0, lim[𝑓(𝑥)●𝑔(𝑥)] = +∞

lim+ 𝑓(𝑥) = +∞

𝑥→𝑎

lim 𝑓(𝑥) = −∞

𝑥→𝑎+

lim 𝑓(𝑥) = +∞ lim− 𝑓(𝑥) = −∞

𝑥→𝑎− 𝑥→𝑎

CONTINUITY A function is said to be continuous at 𝑥 = 𝑎 if there is no interruption in the graph of f (x) at a. Its graph is unbroken at a, and there is no hole, jump, or gap. a. CONTINUITY AT A POINT

A function is said to be continuous at 𝑥 = 𝑎 if

the following three conditions are satisfied:

(i) (ii) (iii)

𝑓(𝑥) is defined, that is, it exists at 𝑥=𝑎 lim 𝑓(𝑥) exists lim 𝑓(𝑥) = 𝑓(𝑎) 𝑥→𝑎

𝑥→𝑎

If one or more of these conditions are not satisfied, we say that the function is discontinuous at 𝑥 = 𝑎.

NOTE: this will still hold true if lim 𝑓(𝑥) is 𝑥→𝑎 replaced by lim+ 𝑓(𝑥) or lim− 𝑓(𝑥) = 𝑏. 𝑥→𝑎

𝑥→𝑎

CONTINUITY ON AN OPEN INTERVAL A function is said to be continuous on an open interval (c, d) if it is continuous at each point I that interval.

CONTINUITY ON A CLOSED INTERVAL A function is said to be continuous on an open interval [a,b] if and only if it is continuous on the open interval (a,b), as well as continuous from the right of a and from the left of b. CONTINUITY FUNCTION

OF

A

COMPOSITE

a. LIMIT OF A COMPOSITE FUNCTION -if lim 𝑔(𝑥) = 𝑏 and if the function f is 𝑥→𝑎

lim (𝑓 ◦ 𝑔)(𝑥) = 𝑥→𝑎

𝑓(𝑏), or, equivalently, lim 𝑓(𝑔(𝑥)) = continuous at b,

𝑓(lim 𝑔(𝑥)) 𝑥→𝑎

𝑥→𝑎

b. If the function g is continuous at a and the function f is continuous at g(a), then the composite function 𝑓 ◦ 𝑔 is continuous at a. REMOVABLE DISCONTINUITY A function is said to have a removable discontinuity at 𝑥 = 𝑎, if (i)

(ii)

lim 𝑓(𝑥)exists, and

𝑥→𝑎

lim 𝑓(𝑥) ≠ 𝑓(𝑎)

𝑥→𝑎

JUMP DISCONTINUITY A function is said to have a jump discontinuity at 𝑥 = 𝑎, if lim− 𝑓(𝑥) ≠ 𝑥→𝑎

lim+ 𝑓(𝑥) for a any positive integer.

𝑥→𝑎

b. THEOREMS

1. If f and g are two functions continuous at the number a, then i. 𝑓 + 𝑔 is continuous at a

ii. 𝑓 − 𝑔 is continuous at a

iii. 𝑓 ● 𝑔 is continuous at a

iv. 𝑓 / 𝑔 is continuous at a, provided that 𝑔 (𝑎) ≠ 0

2. A polynomial function is continuous at every number

3. A rational function is continuous at every

𝑛 4. If n is a positive integer and 𝑓(𝑥) = √ 𝑥,

number in its domain. then

•

(i)

If n is odd f is continuous at every number

(ii)

If n is even, f is continuous at every positive number

LIMITS OF TRANSCENDENTAL FUNCTIONS a. EXPONENTIAL FUNCTIONS 1. LIMIT of 𝑦 = 𝑎 > 1 is (i) lim+ 𝑎 𝑥 = 1 (ii)

𝑥→0

lim 𝑎 𝑥 = 1

𝑥→0−

lim 𝑎 = +∞

(iv)

𝑥→+∞

(ii)

𝑥→0

(iii)

𝑎𝑥 , 𝑥

lim 𝑎 𝑥 = 0

𝑥→−∞

2. LIMIT of 𝑦 = 𝑎 𝑥 , 0 < 𝑎 < 1 is (i) lim+ 𝑎 𝑥 = 1 (iii) (iv)

lim 𝑎 𝑥 = 1

𝑥→0−

lim 𝑎 𝑥 = 0

𝑥→+∞

lim 𝑎 𝑥 = +∞

𝑥→−∞

3. LIMIT of 𝑦 = 𝑒 𝑥 is (i) lim+ 𝑒 𝑥 = 1 (ii)

(iii) (iv)

𝑥→0

lim− 𝑒 𝑥 = 1

𝑥→0

lim 𝑒 𝑥 = +∞

𝑥→+∞

lim 𝑒 𝑥 = 0

𝑥→−∞

4. LIMIT of 𝑦 = 𝑒 −𝑥 is (i) lim 𝑒 −𝑥 = 1 𝑥→0 (ii) lim+ 𝑒 −𝑥 = 1 − 𝑥→0 (iii) lim 𝑒 −𝑥 = 0 (iv)

𝑥→+∞ −𝑥 lim 𝑒 𝑥→−∞

= +∞

b. LOGARITHMIC FUNCTIONS LIMIT OF 𝑦 = 𝑙𝑛𝑥 (i) lim 𝑙𝑛𝑥 = −∞ 𝑥→0+

(ii) lim 𝑙𝑛𝑥 = +∞ 𝑥→+∞

c. TRIGONOMETRIC FUNCTIONS (i) lim sin 𝜃 = 0 (ii)

(iii) (iv)

𝜃→0

lim cos 𝜃 = 1

𝜃→0

lim

𝜃→0

lim

𝜃→0

1−cos 𝜃 𝜃

sin 𝜃

𝜃

=0

=1

d. HYPERBOLIC FUNCTIONS 1. HYPERBOLIC SINE (i) lim+ sinh 𝑥 = 0 (ii)

(iii) (iv)

𝑥→0

lim sinh 𝑥 = 0

𝑥→0−

lim sinh 𝑥 = +∞

𝑥→+∞

lim sinh 𝑥 = −∞

𝑥→−∞

2. HYPERBOLIC COSINE (i) lim+ cosh 𝑥 = 1 (ii)

(iii) (iv)

𝑥→0

lim cosh 𝑥 = 1

𝑥→0−

lim cosh 𝑥 = +∞

𝑥→+∞

lim cosh 𝑥 = +∞

𝑥→−∞

3. HYPERBOLIC TANGENT (i) lim+ tanh 𝑥 = 0 (ii)

(iii) (iv)

𝑥→0

lim tanh 𝑥 = 0

𝑥→0−

lim tanh 𝑥 = 1

𝑥→+∞

lim tanh 𝑥 = −1

𝑥→−∞

A. Derivatives of A Function 𝑑𝑦 ) − 𝑓(𝑥1 ) ) 𝑚 = 𝑓 ′ (𝑥) = 𝑑𝑥 = lim 𝑓(𝑥1 + ∆𝑥∆𝑥 ( ∆𝑥→0 B. Basic Derivation Formulas

1. 2.

dx

2. Addition: d [ f ( x) + g ( x )] = d f ( x) + d g (x ) dx dx dx

3.

3. Subtraction:

4.

d d d [ f ( x ) − g ( x )] = f ( x ) − g (x ) dx dx dx

5.

4. Function with A Constant: d d [c f (x )] = c  f ( x) dx dx

6.

 

d x e = ex dx

6. Natural Logarithmic Rule:

d ln x  = 1 dx x

8. Product Rule: d  f (x ) g ( x )  = f ( x ) g ( x) + g ( x) f ( x) dx

3.

1. 2. 3. 4. 5.

𝑑 (𝑢𝑣) = 𝑑𝑥 𝑑(𝑢𝑣𝑤) 𝑑𝑥

𝑢

𝑑( ) 𝑣 𝑑𝑥

𝑑𝑢 𝑛 𝑑𝑥

𝑑√𝑢 𝑑𝑥

=

𝑢

𝑑𝑉

𝑑𝑥

= 𝑢𝑣 𝑑𝑢

𝑑𝑊 𝑑𝑥

𝑑𝑉

𝑉 𝑑𝑥−𝑈 𝑑𝑥 𝑉2

= 𝑛𝑢𝑛−1 =

+𝑉

𝑑𝑢/𝑑𝑥 2√𝑢

𝑑𝑢

𝑑𝑥

𝑑𝑈 𝑑𝑥

+ 𝑢𝑤

𝑑𝑉

𝑑𝑥

+ 𝑣𝑤

𝑑𝑈 𝑑𝑥

𝑑(𝑡𝑎𝑛 𝑢) 𝑑𝑥

𝑑(𝑐𝑜𝑡 𝑢) 𝑑𝑥

𝑑(𝑠𝑒𝑐 𝑢) 𝑑𝑥

𝑑𝑢 𝑑𝑥

= − sin 𝑢

𝑑𝑢

= − sec2 𝑢

= − csc2 𝑢

𝑑𝑥

𝑑𝑢

𝑑𝑥

𝑑𝑢

𝑑𝑥

= − sec 𝑢 tan 𝑢

𝑑(𝑐𝑠𝑐 𝑢) = 𝑑𝑥

𝑑(𝑎𝑟𝑐 cos 𝑢) 𝑑𝑥

𝑑(𝑎𝑟𝑐 tan 𝑢) 𝑑𝑥

𝑑(𝑎𝑟𝑐 cot 𝑢) 𝑑𝑥

𝑑(𝑎𝑟𝑐 sec 𝑢) 𝑑𝑥

𝑑(𝑎𝑟𝑐 csc 𝑢) 𝑑𝑥

6.

C. Derivatives of Algebraic Functions

𝑑𝑥

𝑑𝑥

4. 5.

d  f ( g ( x) = f ( g ( x)) g ( x) dx

= cos 𝑢

𝑑𝑥 𝑑(𝐶𝑜𝑠 𝑢)

𝑑(𝑎𝑟𝑐 sin 𝑢)

1. 2.

10. Chain rule:

𝑑(𝑆𝑖𝑛 𝑢)

𝑑𝑥

− csc 𝑢 cot 𝑢

𝑑𝑢

𝑑𝑥

𝑑𝑢

𝑑𝑥

E. Derivatives of Inverse Trigonometric Functions

7. Logarithmic Rule: d loga x  = 1 x ln(a ) dx

9. Quotient Rule: d  f ( x)  g( x) f  ( x) − f ( x) g ( x) = dx  g( x)   g (x )2

𝑑𝑥 𝑑𝑢2 𝑉

𝑢

= D. Derivatives of Trigonometric Functions 6.

1. Constant Function: d (c ) = 0

5. Natural Exponential Rule:

𝑑( 𝑣 )

𝑑𝑢 𝑑𝑥

=

√1−𝑢2 𝑑𝑢 𝑑𝑥

−

=

√1−𝑢2 𝑑𝑢 𝑑𝑥

=

1+𝑢2

=− =

=−

𝑑𝑢 𝑑𝑥

1+𝑢2

𝑑𝑢 𝑑𝑥 𝑢√𝑢2 −1

𝑑𝑢/𝑑𝑥

𝑢√𝑢 2 −1

F. Derivatives of Logarithmic Functions 1. 2. 3.

𝑑(log𝑎 𝑢) 𝑑𝑥

𝑑(ln 𝑢) 𝑑𝑥

=

𝑑(log10 𝑢) 𝑑𝑥

= log 𝑎 𝑒 𝑑𝑢 𝑑𝑥

𝑢

𝑑𝑢 𝑑𝑥

𝑢

= log10 𝑒

𝑑𝑢/𝑑𝑥 𝑢

NOTE: 𝑙𝑜𝑔10 𝑒 = 0.43429

G. Derivatives of Exponential Functions 1. 2.

𝑑(𝑎 𝑢 ) 𝑑𝑥

𝑑(𝑒 𝑢 ) 𝑑𝑥

= 𝑎 𝑢 ln 𝑎

= 𝑒𝑢

𝑑𝑢 𝑑𝑥

𝑑𝑢

𝑑𝑥

3. 4.

𝑑(𝑢𝑣 )

= 𝑣𝑢𝑣−1

𝑑𝑥 𝑎 𝑢) 𝑑(log 𝑑𝑥

=

+𝑢𝑣 ln 𝑢

𝑑𝑢

𝑑𝑥 𝑑𝑥 log𝑎 𝑒 𝑑𝑢 𝑢

𝑑𝑥 𝑑𝑣

H. Derivatives of Hyperbolic Functions 1. 2. 3. 4. 5. 6.

= cos ℎ 𝑢

𝑑(𝑠𝑖𝑛 ℎ 𝑢) 𝑑𝑥

= sin 𝑢

𝑑(cos ℎ 𝑢) 𝑑𝑥

𝑑𝑥

𝑑𝑥

𝑑𝑥

𝑑𝑢

𝑑𝑥

= − csc ℎ2 𝑢

𝑑(cot ℎ 𝑢)

𝑑(csc ℎ 𝑢)

𝑑𝑥

𝑑𝑥

= sec ℎ 2 𝑢

𝑑(tan ℎ 𝑢)

𝑑(sec ℎ 𝑢) 𝑑𝑥

𝑑𝑢

𝑑𝑢

𝑑𝑢

𝑑𝑥

= −sec ℎ 𝑢 tan ℎ 𝑢

= −csc ℎ 𝑢 cot ℎ 𝑢

𝑑𝑢

𝑑𝑥

𝑑𝑢

𝑑𝑥

I. Derivatives of Inverse Hyperbolic Functions 1. 2. 3. 4. 5. 6. 7.

𝑑𝑥

=

𝑑(tan ℎ −1 𝑢) 𝑑𝑥

=

𝑑(sec ℎ −1 𝑢)

=

𝑑(sin ℎ −1 𝑢)

𝑑(cos ℎ−1 𝑢) 𝑑𝑥

𝑑(cot ℎ −1 𝑢) 𝑑𝑥

𝑑𝑥

𝑑(csc ℎ−1 𝑢) 𝑑𝑥

𝑑(𝑔 𝑑𝑢) 𝑑𝑥

= =

=

𝑑𝑢 𝑑𝑥

∙ sin ℎ−1 𝑢

√1+𝑢2

𝑑𝑢 𝑑𝑥 √𝑢2 −1 𝑑𝑢 𝑑𝑥

1+𝑢2 𝑑𝑢 𝑑𝑥

1−𝑢2 𝑑𝑢 𝑑𝑥

∙ cos ℎ −1 𝑢

∙ tan ℎ −1 𝑢 ∙ cot ℎ −1 u

𝑢√1−𝑢 2 𝑑𝑢 𝑑𝑥

𝑢√1+𝑢2

= sec ℎ 𝑢 𝑑𝑥 𝑑𝑢

∙ sec ℎ−1 𝑢

∙ csc ℎ−1 𝑢

Sample Problems & Solutions

4) The kinetic energy of K of a body of mass m and velocity v is given by

EASY 1) Given the √𝑥𝑦 = 𝑥 + 𝑦, where y is an implicit function of x, solve for y’

Solution: √𝑥𝑦 = 𝑦 + 𝑥

1 1 (𝑥𝑦)−2 (𝑥𝑦′ + 𝑦) = 𝑦′ + 1 2

𝑥𝑦′ + 𝑦 2√𝑥𝑦

𝐾 = 𝑚𝑣 2 2 1

(a) Solve for

Solution: 𝐾 = 2 𝑚𝑣 2 ; 1

2) If 𝑦 = 2 (2𝑥 2 + 𝑥) 1

2√𝑥𝑦 − 𝑦

, find

Solutions: 𝑦 = 2 (2𝑥 2 + 𝑥)−3 ; 1

𝑑𝑦

𝑑𝑥

3 = − (2𝑥 2 + 𝑥)−4 (4𝑥 + 1) 2 3 4𝑥 + 1 𝑑𝑦 =− [ ] 𝑑𝑥 2 (2𝑥 2 + 𝑥) 4

Solutions: 𝑦 = 3𝑥 2 − 2𝑥 + 6

𝑑𝑦 = (6𝑥 − 2)𝑑𝑥

𝑤ℎ𝑒𝑟𝑒 𝑑𝑥 = 2.03 − 2 = 0.03 ; 𝑥 = 2 𝑑𝑦 = [6(2) − 2](0.03) 𝑑𝑦 = 0.3

𝜕𝑣 2

=𝐾

1 2 𝑣 ⋅𝑚=𝑘 2 1 𝑚𝑣 2 = 𝑘 2

5) Find the slope of the tangent to the curv...