# 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**

## 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.

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