Calculus
Anton Bivens Davis
10 TH Edition
Shahbaz Ahmed
email:
Ex 2.8
Implicit Differentiation
If the reader find any answer uncomfortable ,he/she may email me and i shall
immediately revise the same answer up to the entire satisfaction of the learner.To
satisfy the learner is my professional commitment and ultimate goal
Q1
dx dy
Equation : y = 3x + 5. (a) Given that dt = 2, find dt when x = 1.
dy dx
(b) Given that dt = −1 , find dt when x = 0
Solution
y = 3x + 5
Taking derivative with respect to t
1
,dy d(3x+5) d(3x) d(5)
dt = dt = dt + dt
dy d(3x) dx d(5) dx
dt = dx dt + dx dt
dy
dt = 3 dx dx
dx dt + 0 ×
dx
dt
dy
dt = 3 dx dx
dt + 0 = 3 dt .................. eq (1)
dx
(a)Putting dt = 2 in equation (1)
dy
dt = 3(2) = 6 when x=1
dy
(b) Putting dt = −1 in equation (1)
−1 = 3 dx
dt
dx
dt = − 31 when x=0
Q2
Equation: x + 4y = 3.
dx dy
(a) Given that dt = 1, find dt when x = 2.
dy dx
(b) Given that dt = 4 , find dt dx/dt when x = 3.
Solution
x + 4y = 3
Taking derivative with respect to t
2
,d(x+4y) d(3)
dt = dt
dx d(4y)
dt + dt =0
dx
dt + 4 dy
dt = 0 ....................... eq (1)
dx
(a) Putting dt = 1 in equation (1)
1 + 4 dy
dt = 0 Or
4 dy
dt = −1 Or
dy
dt = − 41
dy
(b)Putting dt = 4 in equation (1)
dx
dt +4×4=0
dx
dt = −16
Q3
Equation: 4x2 + 9y 2 = 1
dx dy
(a) Given that dt = 3, find dt when
1 1
(x, y) = ( 2√ , √
2 3 2
)
dy dx
(b) Given that dt = 8, find dt when
√
5
(x, y) = ( 13 , − 9 ) Solution
3
, 4x2 + 9y 2 = 1
Taking derivative with respect to t
d(4x2 +9y 2 )) d(1)
dt = dt
d(4x2 ) d(9y 2 )
dt + dt =0
d(4x2 ) dx d(9y 2 ) dy
dx dt + dy dt =0
2 2
4 d(x ) dx d(y ) dy
dx dt + 9 dy dt = 0
dy
4 × 2x dx
dt + 9 × 2y dt = 0
dy
8x dx
dt + 18y dt = 0 ........................ eq (1)
dx 1 1
(a)Putting dt = 3, x = √
2 2
,y = √
3 2
in equation (1)
1 1 dy
8× √
2 2
× 3 + 18 × √
3 2 dt
=0
1 dy 1
18 × √
3 2 dt
= −8 × √
2 2
×3
dy 1
√ 1
dt = −8 × √
2 2
×3×3 2× 18 = −2
√
dy 5
(b) Putting dt = 8, x = 13 , y = − 9 in equation (1)
√
1 dx 5
8× 3 dt − 18 × 9 ×8=0
dx
√
5 1
√
dt = 18 × 9 ×8×3× 8 =6 5
Q4
4
Anton Bivens Davis
10 TH Edition
Shahbaz Ahmed
email:
Ex 2.8
Implicit Differentiation
If the reader find any answer uncomfortable ,he/she may email me and i shall
immediately revise the same answer up to the entire satisfaction of the learner.To
satisfy the learner is my professional commitment and ultimate goal
Q1
dx dy
Equation : y = 3x + 5. (a) Given that dt = 2, find dt when x = 1.
dy dx
(b) Given that dt = −1 , find dt when x = 0
Solution
y = 3x + 5
Taking derivative with respect to t
1
,dy d(3x+5) d(3x) d(5)
dt = dt = dt + dt
dy d(3x) dx d(5) dx
dt = dx dt + dx dt
dy
dt = 3 dx dx
dx dt + 0 ×
dx
dt
dy
dt = 3 dx dx
dt + 0 = 3 dt .................. eq (1)
dx
(a)Putting dt = 2 in equation (1)
dy
dt = 3(2) = 6 when x=1
dy
(b) Putting dt = −1 in equation (1)
−1 = 3 dx
dt
dx
dt = − 31 when x=0
Q2
Equation: x + 4y = 3.
dx dy
(a) Given that dt = 1, find dt when x = 2.
dy dx
(b) Given that dt = 4 , find dt dx/dt when x = 3.
Solution
x + 4y = 3
Taking derivative with respect to t
2
,d(x+4y) d(3)
dt = dt
dx d(4y)
dt + dt =0
dx
dt + 4 dy
dt = 0 ....................... eq (1)
dx
(a) Putting dt = 1 in equation (1)
1 + 4 dy
dt = 0 Or
4 dy
dt = −1 Or
dy
dt = − 41
dy
(b)Putting dt = 4 in equation (1)
dx
dt +4×4=0
dx
dt = −16
Q3
Equation: 4x2 + 9y 2 = 1
dx dy
(a) Given that dt = 3, find dt when
1 1
(x, y) = ( 2√ , √
2 3 2
)
dy dx
(b) Given that dt = 8, find dt when
√
5
(x, y) = ( 13 , − 9 ) Solution
3
, 4x2 + 9y 2 = 1
Taking derivative with respect to t
d(4x2 +9y 2 )) d(1)
dt = dt
d(4x2 ) d(9y 2 )
dt + dt =0
d(4x2 ) dx d(9y 2 ) dy
dx dt + dy dt =0
2 2
4 d(x ) dx d(y ) dy
dx dt + 9 dy dt = 0
dy
4 × 2x dx
dt + 9 × 2y dt = 0
dy
8x dx
dt + 18y dt = 0 ........................ eq (1)
dx 1 1
(a)Putting dt = 3, x = √
2 2
,y = √
3 2
in equation (1)
1 1 dy
8× √
2 2
× 3 + 18 × √
3 2 dt
=0
1 dy 1
18 × √
3 2 dt
= −8 × √
2 2
×3
dy 1
√ 1
dt = −8 × √
2 2
×3×3 2× 18 = −2
√
dy 5
(b) Putting dt = 8, x = 13 , y = − 9 in equation (1)
√
1 dx 5
8× 3 dt − 18 × 9 ×8=0
dx
√
5 1
√
dt = 18 × 9 ×8×3× 8 =6 5
Q4
4
