Suppose y= √2x + 1 where x and y are functions of t.
(a) if dx/dt= 12, find dy/dt when x=4.
(b) if dy/dt= 3, find dx/dt when x=4012. 

Solution:

y = √2x + 1 [Given]

Differentiating both sides with respect to t

dy/dt = dx/dt/ √(2x + 1)

(a) if dx/dt= 12, we have to find dy/dt when x=4

dy/dt = 12/√(2(4) + 1)

dy/dt = 12/√(9)

dy/dt = 12/√9

dy/dt = 12/3

dy/dt = 4

(b) if dy/dt= 3, we have to find dx/dt when x=4012. 

3 = dx/dt/ √(2(4012) + 1)

3 × √(8024 + 1) = dx/dt

dx/dt = 3 × √8025

dx/dt = 3 × 5√321

dx/dt = 15 √321

Therefore, the values of dy/dt is 4 and dx/dt is 15√321.

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Suppose y= where x and y are functions of t.
(a) if dx/dt= 12, find dy/dt when x=4.
(b) if dy/dt= 3, find dx/dt when x=4012. 

Summary:

Suppose √2x + 1 where x and y are functions of t.

(a) if dx/dt= 12, dy/dt is 4 when x = 4.

(b) if dy/dt= 3, dx/dt is 15√321 when x = 4012.