Question

I answered this problem wrong on my Calculus BC test. Can you guys help me figure out the correct answer?

If y = 3x^3 - 2x and dx/dt = 3 , find dy/dt when x = -2.

Answer 1

`(dy)/(dt)=102`

Explanation:

It was given that;

`y=3x^3-2x`

When we differentiate this function with respect to x, we get;

`(dy)/(dx)=9x^2-2`

When
`x=-2`, we get;

`(dy)/(dx)=9(-2)^2-2`

`(dy)/(dx)=34`

Also we were given that;

`(dx)/(dt) =3`.

Recall the chain rule;

`(dy)/(dx)=(dy)/(dt) * (dt)/(dx)`

`\Rightarrow (dy)/(dt)=(dy)/(dx) * (dx)/(dt)`

We substitute these values into the formula to get;

`\Rightarrow (dy)/(dt)=34* 3`

`(dy)/(dt)=102`