Suppose y = √(2x + 1), where x and y are functions of t.
a)if dx/dt = 3, find dy/dt when x = 4
b)if dy/dt = 5, find dx/dt when x = 12

Solution:

We will differentiate the given parametric 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 (3) + 0

dy/ dt = 1/ 2 (9)-1/2 × 6

dy/ dt = 1/ 91/2 × 3

dy/ dt = 1/ 3 × 3

dy/ dt = 1

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 = 5 and x = 12.

5 = 1/ 2 (2(12) + 1)^-1/2 × 2 dx/ dt + 0

5 = (24 + 1)^-1/2 dx/ dt

5 = 5^-1/2 × dx/ dt

5 = 1/ 5 × dx/ dt

dx/ dt = 25

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Suppose y = √(2x + 1), where x and y are functions of t.
a)if dx/dt = 3, find dy/dt when x = 4
b)if dy/dt = 5, find dx/dt when x = 12

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.