# if y= sqrt(2x^2-2), determine dy/dt when x=3, dx/dt=-4 dy/dt=

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**Sorry about the typo in the previous response.**

The function `y = sqrt(2x^2 - 2)` . Take the derivative with respect to t.

`dy/dt = (4x)/(2*sqrt(2x^2 - 2))*(dx/dt)`

=> `dy/dt = (2x)/sqrt(2x^2 - 2)*(dx/dt)`

When x = 3, `y = +- 4` . As `dx/dt = -4` ,

`dy/dt = 6/4*-4 = -6` or `dy/dt = -6/4*-4 = 6`

There are two points at which the given conditions are satisfied.

**The value of `dy/dt` can be either 6 or -6**

The function `y = sqrt(2x^2 - 2)` . Take the derivative with respect to t.

`dy/dt = (4x)/(2*sqrt(2x^2 - 2))*(dx/dt)`

=> `dy/dt = (2x)/(sqrt(2x^2 - 2))*(dx/dt)`

When x = 3 and `dx/dt = -4`

`dy/dt = 6/sqrt 16*-4`

=> `dy/dt = +- 3/2 - 4`

=> `dy/dt = -11/2` and `dy/dt = -5/2`

There are two points at which the given conditions are satisfied.

**The value of `dy/dt` can be eitherĀ ****`-11/2` orĀ `-5/2` **