Suppose y = √(2x + 1), where x and y are functions of t
(a) If dx/dt = 15, find dy/dt, when x = 4
(b) If dy/dt = 2, find dx/dt, when x = 40

Solution:

We will differentiate the given function with respect to t using the chain rule and substitute the given values.

a) Given, y = √(2x + 1) can be written as y = (2x + 1)^1/2

Differentiate the function w. r. to t.

dy/ dt = 1/ 2 (2x + 1)^-1/2 × 2 dx/ dt + 0

Substitute the values of dx/dt = 15 and x = 4.

dy/ dt = 1/ 2 (2(4) + 1)^-1/2 × 2 (15) + 0

dy/ dt = 1/ 2 (9)^-1/2 × 30

dy/ dt = 1/ 9^1/2 × 15

dy/ dt = 1/ 3 × 15

dy/ dt = 5

b) Given, y = √(2x + 1) can be written as y = (2x + 1)^1/2

Differentiate the function w. r. to t.

dy/ dt = 1/ 2 (2x + 1)^-1/2 × 2 dx/ dt + 0

Substitute the values of dy/dt = 2 and x = 40.

2 = 1/ 2 (2(40) + 1)^-1/2 × 2 dx/ dt + 0

2 = 1/ 2 (81)^-1/2 × 2 dx/ dt

2 = 1/ 81^1/2 × dx/ dt

2 = 1/ 9 × dx/ dt

dx/ dt = 18

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Suppose y = √(2x + 1), where x and y are functions of t
(a) If dx/dt = 15, find dy/dt, when x = 4
(b) If dy/dt = 2, find dx/dt, when x = 40

Summary:

The value of dy/dx = 5 when dx/dt = 15,and x = 4. Also, dx/ dt = 18 when dy/dt = 2 and x = 40.