UNIT 3 MODULE 5 final.pdf

UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION ALGEBRAIC FUNCTION An algebraic function is one formed by a finite number of algebraic operations on constants and/or variables. These algebraic operations include addition, subtraction, multiplication, division, raising to powers, and extracting ro...

UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION ALGEBRAIC FUNCTION An algebraic function is one formed by a finite number of algebraic operations on constants and/or variables. These algebraic operations include addition, subtraction, multiplication, division, raising to powers, and extracting roots. Polynomial and rational functions are particular kinds of algebraic functions. Polynomial function is defined by f x   an x n  an1 x n1  an2 x n2 ...  a1 x  a0  Where a0 , a1 ,..., an are real numbers a n  0 and  is a non-negative integer. Function f is a polynomial function of degree n. Rational function is a function that can be expressed as a quotient of two polynomial functions. x 1 x 2  25 Example 1. f x   2 Example 2. hx   x 4 x3 DIFFERENTIATION FORMULAS OF ALGEBRAIC FUNCTIONS The following differentiation formulas were derived using the Increment Method of differentiation. On the list below, let and be functions of ; while and are constants. 1. Derivative of a Constant 6. Derivative of Square Root of a Function d c   0 d u 1 d u  dx dx 2 u dx 2. Derivative of x with respect to x 7. Derivative of a Product of Two Factors d x   1 d uv   u d v   v d u  dx dx dx dx 3. Derivative of a Power of x 8. Derivative of a Product of Three Factors d dx cx n   cnx n1 d dx uvw  uv d w  uw d v  vw d u dx dx dx 4. Derivative of a Sum/Difference of Terms 9. Derivative of a Quotient d u  v   d u   d v  d vu   u v  d dx dx dx d u    dx dx dx  v  v 2 5. Derivative of a General Power 10. Derivative of a Constant Over a Function d cu n   cnu n1 u  d d  c  c d   u  dx dx dx  u  u 2 dx Let me illustrate how to use the listed differentiation formulas. Most people find the differentiation process hard. Students failed to arrive at the correct derivative because of inadequate knowledge of trigonometry, geometry and algebra. But do not be threatened, illustrative examples have steps presented in detailed way. Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 59 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION or f ' x  of the given algebraic functions. dy Example 1. Find the derivative dx a. y  x3  4 x 2  6 x  8 The given function is a sum of terms. To differentiate, we use Formula 4: d u  v   d u   d v  , followed with the use of Formula 3: d cxn   cnxn1 as well as dx dx dx dx Formula 2: d x   1 and Formula 1: c   0. d dx dx ( ) ( ) ( ) (Recall: )  b. y  2 x 2  3  2 There are two possible ways of finding the derivative of the given function which is a general power of. Method 1. First, we transform the given function to a sum of terms by expanding the right side of the given equation above using the special product called square of a binomial: a  b   a 2  2ab  b 2. Hence, 2   2 y  4 x 4  12 x 2  9  4 x 4  12 x 2  9  4 x 4  12 x 2  9  44x 4 1  122x 2 1  0 dy d d d dx dx dx dx dy dx   16 x3  24 x  8 x 2 x 2  3  Method 2. We use directly the Formula 5: d dx   cu n  cnu n1 u  , with d dx u  2 x 2  3, n  2 dy   2 2x2  3  2 1 d    d  2 x2  3  2 2 x2  3  2 x2  3 d  dx dx  dx dx  dy dx    2 2 x 2  3 22x 2 1  0   dy dx      2 2 x 2  3 4 x   8 x 2 x 2  3 Observe that both methods of finding the derivative give to same result. Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 60 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION c. y  4 x  13 x2 This time, the given function is a quotient having the numerator a special product called a cube of binomial. Method 1. First, we expand the special product a  b  a3  3a 2b  3ab2  b3 , the result of 3 which, then divided by x 2. Doing so results to: y 4 x   34 x  1  34 x 1  1 3 2 2 3 x2 64 x3  48 x 2  12 x  1 y  64 x  48  12 x 1  x  2 x2 The simplified form is now a sum of terms and we differentiate using Formula 4: d u  v   d u   d v . dx dx dx  64 x   48  dy d dx dx d dx d dx  12 x 1   d 2 dx x    641x 0  0  12 1x  2   2x 3 dy dx 𝑛 dy 12 2 Recall: 𝑛 𝑎. Hence: 𝑥  64  2  3 𝑎 𝑥3 dx x x v u   u v  d d d u Method 2. We use directly the quotient Formula 9:    dx dx , with: dx  v  v 2 ( ) and. We need and for substitution on the formula. u  4 x  1 𝑣 𝑥 3 du dx  34 x  1 2 d dx  4 x  1  34 x  12 4 x0  0  𝑑𝑣 𝑑𝑥 𝑥  34 x  1 4  124 x  1 du 2 2 dx Substitution on the quotient formula gives y'   2  x 2 124 x  1  4 x  1 2 x  3 x  2 2 Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 61 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION Simplify the numerator by bringing-out the common monomial factor ( ). 2 x4 x  1 6 x  4 x  1 2 x4 x  1 6 x  4 x  1 2 2 y'   x4 x4 y'  2    2 x4 x  1 2 x  1 2 x 16 x 2  8 x  1 2 x  1 x4 x4 y'   2 32 x3  16 x 2  2 x  16 x 2  8 x  1 x3 64 x3  12 x  2 64 x 3 12 x 2 y'   3  3  3 x3 x x x 12 2 y '  64  2  3 x x Note: The derivative was expanded to show that the result of Method 2 is right. However, it is always best to express the derivative of a function in its factored- form.   d. y  3x 4  2 x 2  4 x  1 x5  4 x  2  Method 1. Get the product of the factors on the right side of the given equation to bring the product to a sum of terms, then, use the Formula 4: u  v   u   v . d d d dx dx dx y  3x 9  2 x 7  4 x 6  13 x5  6 x 4  8 x 3  20 x 2  12 x  2 dy d 9 d d d d d d d d  3x  2 x 7  4 x 6  13 x 5  6 x 4  8 x 3  20 x 2  12 x  2 dx dx dx dx dx dx dx dx dx dx  39x8  27 x 6  46x 5  135x 4  64x 3  83x 2  202x  12 dy dx dy  27 x8  14 x 6  24 x 5  65 x 4  24 x 3  24 x 2  40 x  12 dx Method 2. Use the product Formula 9: d uv   u d v   v d u  with: dx dx dx u  3x 4  2 x 2  4 x  1 du  12 x 3  4 x  4 dx Substitute now the above on Formula 9. Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 62 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION dy dx      3x 4  2 x 2  4 x  1 5 x 4  4  x 5  4 x  2 12 x 3  4 x  4   dy  15 x8  10 x 6  20 x5  5 x 4  12 x 4  8 x 2  16 x  4 dx   12 x8  48 x 4  24 x 3  4 x 6  16 x 2  8 x  4 x 5  16 x  8  dy  27 x8  14 x 6  24 x 5  65 x 4  24 x 3  24 x 2  40 x  12 dx dy  27 x8  14 x 6  24 x 5  65 x 4  24 x 3  24 x 2  40 x  12 dx e. y  x 3  4 x 2  6 x  7 Solution: To differentiate, we use the Formula 6: d u 1 d u  with: dx 2 u dx du u  x3  4 x 2  6 x  7 ,  3x 2  8 x  6. Substitution of these on Formula 6 results to dx y'  1d 3 x  4x2  6x  7   2 x  4x  6x  7 3 2 dx y'  1 3x 2  8 x  6   2 x  4x  6x  7 3 2 3x 2  8 x  6 y'  2 x3  4 x 2  6 x  7 y'  3x 2   8x  6 x3  4 x 2  6 x  7 Rationalize the fraction.  2 x3  4 x 2  6 x  7  4 f. y  x  25 2 The differentiation of the given function can be dome in three ways. v u   u v  d d d u Method 1. We apply the Quotient Formula 9:    dx dx , with: dx  v  v2 u4 v  x 2  25 du , dv , 0  2x dx dx Substitution yields y '  x 2   25 0   42 x    8x x 2  25  2 x 2  25  2 Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 63 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION d  c  c d Method 2. We use the special Quotient Formula 10:   u  ,with: dx  u  u 2 dx du c  4, u  x 2  25,  2x. dx 4 y'   4 d 2 x  25   2 x    x  25 dx 2  2 x  25 2 2    8x y '  x 2  25  2 Method 3. We may rewrite the given function to y  4x 2  25 , then, use the general 1 power Formula 5: d dx cu n   cnu n1 u  , with c  4, n  1, n  1  2, u  x 2  25, d dx du dx  2x  8x  y '  4 1 x 2  25  2 x   2 x 2  25  2 Observe that all methods of differentiation presented yield same result. 2x g. y  4x  5 The given function can be differentiated in two different ways. Method 1: Rewrite the given function to y  2 x4 x  5 2 , then, apply Product Formula 7: 1  d uv   u d v   v d u . dx dx dx v  4 x  5 2 1  u  2x   4 x  5 2 4 x  5 dv 1  1 d 1 du 2 dx 2 dx dx   4 x  5 2 4   dv 1  3 2 4 x  52 3 dx 2 Therefore, after doing the necessary substitutions, we got  2    4 x  5 2 2 1 y'  2 x   4 x  52  3    4x 2  4x 2 y'     4 x  5 4 x  5 4 x  5 4 x  5 4x  5 3 2 Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 64 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION  4 x  24 x  5  4 x  8 x  10 4 x  10 y'    4 x  5 4 x  5 4 x  5 4 x  5 4 x  5 4 x  5 22 x  5 y'  4 x  5 4 x  5 v u   u v  d d d u Method 2. We use the Quotient Formula 9:    dx dx dx  v  v2 v  4x  5 dv  1 4  2 dx 2 4 x  5 4x  5  2  4 x  5 2  2 x  u  2x  4x  5  y'  4x  5 du 2 24 x  5  4 x 4 x  10 dx y'   4 x  5 4 x  5 4 x  5 4 x  5 22 x  5 y'  4 x  5 4 x  5 4  x3  8  h. y   3   2x 1  Solution: To differentiate the given function, apply the general power Formula 5: cu n   cnu n1 u  , with: d d 3 3. dx dx Using the quotient formula to find      du 2 x 3  1 3x 2  x 3  8 6 x 2     51x 2. dx  2x3 1 2  2x3 1 2   dy  x3  8   4 3  3   51x 2   3    204 x 2 x3  8  3 Therefore, using Formula 5 yields: 2 dx  2x  1     3   2 x  1  2 x 3  1 2 x 3  1 2   dy  204 x 2 x3  8   3 dx  2 x3  1 5  Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 65 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION SAQ8 ACTIVITY 3.5 – A NAME: ____________________________________________________ SCORE: ______________ SECTION: ___________DATE: _______________ PROF: __________________________________ Find the derivative of the following functions. If ever possible, express final answer in factored form. 1. y  3x 5  3x 4  7 x 3  5 x 2  x  8 4  9x 4. y  4  9x 5 3 2. y  3x5  4 x3   x3  2  x2 5. y   3   x 1  3. ( ) 1 6. y  3x  3 5x Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 66 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION ASAQ8 ACTIVITY 3.5 – A NAME: ____________________________________________________ SCORE: ______________ SECTION: ___________DATE: _______________ PROF: __________________________________ Find the derivative of the following functions. If ever possible, express final answer in factored form. 1. y  3x  3x  7 x  5x  x  8 5 4 3 2 4  9x 4. y  4  9x ( ) 5 3 2. y  2 x 5  4 x 3   x3  2  x2 5. y   3   x 1  ( ) ( ) 3. ( ) 6. y  3x  3 5 x ( ) ( ) 3 √ √ Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 67 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION ACTIVITY 3.5 – B NAME: ____________________________________________________ SCORE: ______________ SECTION: ___________DATE: _______________ PROF: __________________________________ Find the derivative of the following functions. If ever possible, express final answer in factored form. 1. y  3x 2  5 x  1 4. y  x5 1  2 x 7  2. y  3x 4  2 x 2  4  2 5. y  3 2 x 2  4 x  3   3. y  x3  6 x 2  4 x3  6. y  2  3x  2 x 2 x3 Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 68 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION 3 x 7. y  10. y  x 2 4  3 2x  1 8. y  x2  1   11. y  x 3  2 3x 2  4 x x4 5 9. y   4   12. y  x 3  2 3x 2  4 5 x Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 69 UNIT 3 – DERIVATIVE OF ALGEBRAIC FUNCTION Differential Calculus Module 5 – Differentiation Formulas for Algebraic Function Page 70