Calculus I (Questions, Detailed Solutions and Answers Practice
Quiz)
1. Find the
rst derivative f 0 (x) of the function f (x) = x2 sin x
Solution: To solve this problem, we use the Product rule :
d(uv) dv du
=u +v
dx dx dx
where u and v are two functions which are dierentiable at x.
For the function f (x) = x2 sin x,
Answer: f (x) = d(x dxsin x) = x d(sin
2 2
0 x) dx 2
+ sin x = x2 cos x + 2x sin x
dx dx
2. If f (x) = x sin x
nd f 0 (x)
Solution: Using the Product rule,
Answer: d(x sin x)
dx
=x
d sin x
dx
+ sin x
dx
dx
= x cos x + sin x
sin x
3. Let f (x) = ,
nd f 0 (x)
cos x
Solution:To solve this problem, we use the Quotient rule :
u du dv
d −u
v
v = dx 2 dx
dx v
sin x
For the function f (x) =
cos x
,
sin x d sin x d cos x
d cos x − sin x
Answer: f (x) =
2 2
0 cos x
= dx dx = cos x + sin x = 1 = sec2 x
dx cos2 x cos2 x cos2 x
sin x
4. If f (x) = ,
nd f 0 (x)
x
Solution: Using the Quotient rule,
sin x d sin x dx
d x − sin x
x dx dx
=
dx x2
Answer :=
x cos x − sin x
x2
5. If f (x) = (1 + x2 )2 ,
nd f 0 (x)
1
, Solution:To solve this problem, we use the Chain rule :
Let f be a function of a variable u, dierentiable at u0 , and let u be a non-constant function of a
variable x dierentiable at x0 with u(x0 ) = u0 . Then the composite function (f ◦ u)(x) ≡ f (u(x)) is
dierentiable at x and is given by,
df df du
= .
dx du dx
For the function f (x) = (1 + x2 )2
df du
Let u = 1 + x2 , then f = u2 , = 2u and = 2x
du dx
Answer:f (x) = dx
df
0
=
df du
du dx
= 2u.2x = 4ux = 4x(1 + x ) 2
dy
6. If y = sin(x2 + 3x − 1)
nd
dx
Solution : Let u = x + 3x − 1,
2 du
dx
= 2x + 3
dy
y = sin u, = cos u
du
Answer :
dy
dx
=
dy du
.
du dx
= cos u.(2x + 3) = cos(x2 + 3x − 1) = (2x + 3) cos(x2 + 3x − 1)
dy
7. Find at the point (x, y) = (2, 1) if x3 + xy + y 5 − 11 = 0
dx
Solution: Unlike problems 1 − 3, in this problem, the function y = f (x) is expressed implicitly
as a function of x. Hence to solve the problem, we will
nd the derivative of each term, while ap-
plying the rules of dierential calculus.
Finding the derivative of each of the terms in x3 + xy + y 5 − 11 = 0, we get,
dy dy
3x2 + y + x 5y 4 = 0, which simpli
es to
dx dx
dy
(x + 5y 4 ) = −(3x2 + y)
dx
dy
Making the subject of the formula, we get:
dx
dy (3x2 + y)
=−
dx x + 5y 4
At the point (x, y) = (2, 1), we have,
Answer: dy
dx
=
−3(2)2 + 1
2+5
=
−13
7
dy
8. Find if xy 5 − x2 y + 5x − 3 = 0
dx
Solution: Applying the chain rule and product rule, we
nd the derivative of each term in
xy 5 − x2 y + 5x − 3 = 0 to get:
dy dy dy
y 5 + 5xy 4 − x2 − 2xy + 5 = 0. This can be simpli
ed to (5xy 4 − x2 ) = 2xy − y 5 − 5.
dx dx dx
2
Quiz)
1. Find the
rst derivative f 0 (x) of the function f (x) = x2 sin x
Solution: To solve this problem, we use the Product rule :
d(uv) dv du
=u +v
dx dx dx
where u and v are two functions which are dierentiable at x.
For the function f (x) = x2 sin x,
Answer: f (x) = d(x dxsin x) = x d(sin
2 2
0 x) dx 2
+ sin x = x2 cos x + 2x sin x
dx dx
2. If f (x) = x sin x
nd f 0 (x)
Solution: Using the Product rule,
Answer: d(x sin x)
dx
=x
d sin x
dx
+ sin x
dx
dx
= x cos x + sin x
sin x
3. Let f (x) = ,
nd f 0 (x)
cos x
Solution:To solve this problem, we use the Quotient rule :
u du dv
d −u
v
v = dx 2 dx
dx v
sin x
For the function f (x) =
cos x
,
sin x d sin x d cos x
d cos x − sin x
Answer: f (x) =
2 2
0 cos x
= dx dx = cos x + sin x = 1 = sec2 x
dx cos2 x cos2 x cos2 x
sin x
4. If f (x) = ,
nd f 0 (x)
x
Solution: Using the Quotient rule,
sin x d sin x dx
d x − sin x
x dx dx
=
dx x2
Answer :=
x cos x − sin x
x2
5. If f (x) = (1 + x2 )2 ,
nd f 0 (x)
1
, Solution:To solve this problem, we use the Chain rule :
Let f be a function of a variable u, dierentiable at u0 , and let u be a non-constant function of a
variable x dierentiable at x0 with u(x0 ) = u0 . Then the composite function (f ◦ u)(x) ≡ f (u(x)) is
dierentiable at x and is given by,
df df du
= .
dx du dx
For the function f (x) = (1 + x2 )2
df du
Let u = 1 + x2 , then f = u2 , = 2u and = 2x
du dx
Answer:f (x) = dx
df
0
=
df du
du dx
= 2u.2x = 4ux = 4x(1 + x ) 2
dy
6. If y = sin(x2 + 3x − 1)
nd
dx
Solution : Let u = x + 3x − 1,
2 du
dx
= 2x + 3
dy
y = sin u, = cos u
du
Answer :
dy
dx
=
dy du
.
du dx
= cos u.(2x + 3) = cos(x2 + 3x − 1) = (2x + 3) cos(x2 + 3x − 1)
dy
7. Find at the point (x, y) = (2, 1) if x3 + xy + y 5 − 11 = 0
dx
Solution: Unlike problems 1 − 3, in this problem, the function y = f (x) is expressed implicitly
as a function of x. Hence to solve the problem, we will
nd the derivative of each term, while ap-
plying the rules of dierential calculus.
Finding the derivative of each of the terms in x3 + xy + y 5 − 11 = 0, we get,
dy dy
3x2 + y + x 5y 4 = 0, which simpli
es to
dx dx
dy
(x + 5y 4 ) = −(3x2 + y)
dx
dy
Making the subject of the formula, we get:
dx
dy (3x2 + y)
=−
dx x + 5y 4
At the point (x, y) = (2, 1), we have,
Answer: dy
dx
=
−3(2)2 + 1
2+5
=
−13
7
dy
8. Find if xy 5 − x2 y + 5x − 3 = 0
dx
Solution: Applying the chain rule and product rule, we
nd the derivative of each term in
xy 5 − x2 y + 5x − 3 = 0 to get:
dy dy dy
y 5 + 5xy 4 − x2 − 2xy + 5 = 0. This can be simpli
ed to (5xy 4 − x2 ) = 2xy − y 5 − 5.
dx dx dx
2
