MATH 22981 Winter 2020: Differential Calculus Assignment 2 Solutions

This document presents the solutions to a Calculus assignment focusing on differential calculus concepts. The assignment includes three main problems. The first problem requires finding the equation of a tangent line to a curve at a given point using the limit definition of the derivative. The second problem involves evaluating the slope of a tangent line using the power law rule. The third problem asks for the evaluation of three different derivatives, applying various differentiation techniques such as the power rule and product rule. The solutions are presented step-by-step, demonstrating the application of calculus principles to solve the problems effectively.

Running head: CALCULUS
CALCULUS
Name of the Student:
Name of the University:
Author Note:

1CALCULUS
Answer to Question 1:
Given,
The equation of the curve:
y = √(1-x) = f(x)
at the point x = 0.
Therefore, f(0) = 1
Point of the tangent is (0,1)
Now the slope of the tangent can be calculated as:
mtan = lim
h→ 0
f ( a+h ) −f ( a)
h [Using limit definition]
= lim
h→ 0
f ( 0+h ) −f (0)
h
= lim
h→ 0
√ 1−h−1
h
Now, Multiplying both numerator and denominator by (√1−h+1) we get,
mtan = lim
h→ 0
( √1−h−1)( √1−h+ 1)
h( √1−h+1)
= lim
h→ 0
( √(1−h) ) 2−12
h ( √ 1−h+1)
= lim
h→ 0
1−h−1
h ¿ ¿ ¿

2CALCULUS
= lim
h→ 0
−h
h ( √1−h+1)
= lim
h→ 0
−1
√1−h+ 1
Putting h=0 in the above equation, we get,
= −1
2
Therefore, the slope of the tangent is −1
2
Now, point is (0,1) and slope is -1/2. Therefore, the equation of the tangent is:
y – 1 = −1
2 (x - 0)
 y -1 = −x
2
 2y -2 = -x
 2y + x -2 = 0
Thus, the equation of the tangent to the curve y = √ (1-x) at x =0 is
2y + x – 2 = 0
Answer to Question 2:
Given curve is:
y = 2
3 x2−1

3CALCULUS
Which can be written as:
y = 2 (3x2 - 1)-1
Using power Law rule, we get:
dy
dx = 2. d
dx ( 1
3 x2−1 )
= -2.
d
dx (3 x2−1)
( 3 x2−1 ) 2
= - 2(3. d
dx [ x2 ] + d
dx [−1])
( 3 x2−1 ) 2
= - 2(3.2 x +0)
( 3 x2−1 )2
= - 12 x
( 3 x2 −1 )
2
Now, slope of the tangent at x =1 is:
mtan at x = 0 is
= - 12 .0
( 3(0)2−1 ) 2 = 0.

4CALCULUS
Answer to Question 3:
1. 𝑦 = 5√ (𝑥2 + 1)
Given:
𝑦 = 5√ (𝑥2 + 1)
Therefore,
dy
dx = 5. dy
dx √ (𝑥2 + 1) [Applying Product rule]
= 5. 1
2 (x2 + 1)1/2 – 1. d
dx x2 + 1 [Applying Power rule]
= 5{ d
dx [ x2 ] + d
dx [ 1 ] }
2(√ (x2+1))
= 5 (2 x +0)
2(√ (x2+1))
= 5 x
√ ( x2+1)¿ ¿
Thus, d
dx (5 √(x2 +1)) = 5 x
√ (x2+1) ¿ ¿
2. 𝑦 = 1/ (∛ 𝑥4 + 2𝑥 – 1)
Given,
𝑦 = 1/ (∛ 𝑥4 + 2𝑥 – 1)
Therefore,

5CALCULUS
dy
dx = -1
3 . (x4 + 2x – 1)-1/3-1. d
dx [x4 + 2x -1] [Applying Power Rule]
= -
d
dx [ x4 ]+2. d
dx [ x ] + d
dx [ −1 ]
3(x4 +2 x – 1)4 / 3
= - 4 x3+2+ 0
3(x4 +2 x – 1)4 / 3 [Applying Power Rule]
= - 4 x3 +2
3( x4 +2 x−1)4 /3
Thus,
d
dx (1/ (∛ 𝑥4 + 2𝑥 – 1)) = - 4 x3 +2
3( x4 +2 x−1)4 /3
3. y = (5 / (3x-2 + 5x))2
Given,
y = ¿
y = 25
(3 x−2 +5 x)2 .
Therefore,
dy
dx = 25. d
dx
1
( 5 x + 3
x2 )
2 [Pulling the constant out from the function]
= 25 (-2) (5 x + 3
x2 )−3
. d
dx [5x + 3
x2 ] (Applying Power rule)

6CALCULUS
= -
50(5. d
dx [ x ] + 3. d
dx [ 1
x2 ])
(5 x+ 3
x2 )3
= -
50(5.1+3 (−2) x−3 )
( 5 x + 3
x2 )
3
= -
50(5− 6
x3 )
( 5 x + 3
x2 )
3
Thus,
d/dx (5 / (3x-2 + 5x))2 = -
50(5− 6
x3 )
(5 x + 3
x2 )3