Questions/Answers on Calculus

Running head: CALCULUS
CALCULUS
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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