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Unit-2: Derivatives | Assignment

Derivative and evaluation questions for Unit 2 of MCV4U course.

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Added on  2022-08-29

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Please not that one is calculus and one is advanced functions and it is clearly indicated in the title.

Unit-2: Derivatives | Assignment

Derivative and evaluation questions for Unit 2 of MCV4U course.

   Added on 2022-08-29

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Running head: UNIT 2: DERIVATIVES
UNIT 2: DERIVATIVES
Name of the Student
Name of the University
Author Note
Unit-2: Derivatives | Assignment_1
UNIT 2: DERIVATIVES1
1.
a. f(x) = 13x^4 – 7x^3 + 5x^2 + 11x+ 75
=> f’(x) = 13*4x^3 + 7*3x^2 + 5*2x + 11
=> f’(x) = 52x^3 + 21x^2 + 10x + 11
b. f(x) = (x^3 + 2x^2 + 4)(x^4-3)
f’(x) = (x^3+2x^2 + 4)(3x^3) + (x^4-3)*(3x^2 + 4x) (by using product rule)
= 6x^6 + 6x^5 + 12x^3 + 3x^6 +4x^5 – 9x^2-12x
= 9x^6 + 10x^5 +12x^3 -9x^2 – 12x
c. f(x) = (3x^2 -6x+7)/(4x-1)
=> f’(x) = ( 4 x1 )( 6 x6 ) ( 3 x26 x+ 7 )4
( 4 x 1 )2 = 24 x230 x+612 x2 +24 x 28
( 4 x1 )2 =
12 x26 x22
( 4 x1 )2
d. f(x) = (4x^3 -7x)^7
=> f’(x) = 7( 4 x3 7 x ) 6
( 12 x27 )
e. 12 = y^2 + x^2
=> 0 = 2y*dy/dx + 2x
=> dy/dx = -x/y = x
12x2
f. f(x) = 156
=> f’(x) = 0
Unit-2: Derivatives | Assignment_2
UNIT 2: DERIVATIVES2
g. f(x) = (1-3x)^2 * (x^2 – 2)^3
f’(x) = 2 ( 13 x )( 3 ) ( x22 ) 3
+32 x ( 13 x ) 2 ( x22 ) 2
= ( 18 x6 ) ( x22 )3
+6 x ( 13 x )2 ( x22 )2
h. f(x) = ( 2 x2 +1
3 x3 +1 )2
=> f’(x) = 2*
2 x2+ 1
3 x3+ 1( 3 x3 +1 ) 4 x+ ( 2 x2 +1 )9 x2
( 3 x3+ 1 ) 2
= ( 4 x2+2)(12 x4 +4 x +18 x4 +9 x2)
( 3 x3 +1 )3 (By
quotient rule)
2.
f(x) = (3x^2-7x+4)^8
f’(x) = 8( 3 x27 x+ 4 )
7
(6 x7)
f’’(x) = 8( 3 x27 x+ 4 )7
6+ 87 ( 3 x27 x + 4 )6
( 6 x7 )( 6 x 7)
= 48 ( 3 x27 x +4 )7
+56 ( 3 x27 x+ 4 )6
( 6 x7 )2
3.
f(x) = 12x^10 – 3x^7 +4x^5 + 5x^2 + 6
f’(x) = 120x^9 – 21x^6 + 20x^4 + 10x
f’’(x) = 108x^8 – 126x^5 + 80x + 10
f’’’(x) = 864x^7 – 630x^4 + 80
4.
y = x^2 -2x + 3
Unit-2: Derivatives | Assignment_3
UNIT 2: DERIVATIVES3
dy/dx = 2x +2
[ dy
dx ](2,3)
=22+2=6 = m
Equation of the tangent line (y-3)/(x-2) = m
(y-3)/(x-2) = 6
y-3 = 6x-12
y = 6x-9
5.
Equation of parabola y = -x^2 + 3x +4
dy/dx = -2x + 3 = m
Hence, m =5 at
-2x + 3 = 5 => x = -1
Hence, y = -1-3+4 = 0
Thus the point where slope of the tangent is 5 is (-1,0)
6.
An example of decreasing rate of velocity of a particle is given by,
d2 x
d t2 =k
The example of increasing rate of velocity of a particle is given by,
d2 x
d t2 =k
where, dx/dt = velocity of particle in m/sec and k>0 and t = time in seconds
Unit-2: Derivatives | Assignment_4

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