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.

Question

asked Oct 24, 2020 48.7k views

1 Answer

BestPractices · Answer 1 · 2020-10-30T18:13:42+0000

Answer:

(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