Question

Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240

dx/dt=

Answer 1

The value of
`(dx)/(dt)` is
`(160)/(x^3)`.

Explanation:

The given equation is

`x^4=8y^5-240`

We need to find the value of
`(dx)/(dt)`.

Differentiate with respect to t.

`4x^3(dx)/(dt)=8(5y^4)(dy)/(dt)-0`
`[\because (d)/(dx)x^n=nx^(n-1),(d)/(dx)C=0]`

`4x^3(dx)/(dt)=40y^4(dy)/(dt)`

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

`4x^3(dx)/(dt)=40(2)^4(1)`

`4x^3(dx)/(dt)=40(16)(1)`

`4x^3(dx)/(dt)=640`

Divide both sides by 4x³.

`(dx)/(dt)=(640)/(4x^3)`

`(dx)/(dt)=(160)/(x^3)`

Therefore the value of
`(dx)/(dt)` is
`(160)/(x^3)`.