Question

find dy/dt where 4x^2-y=100 and dx/dt equals 8 when x=15

Answer 1

use the chain run

\[\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\]

Answer 2

its related rates and change the equation to

\[y = 4x^2 - 100\]

so find dy/dx

Answer 3

Just differentiate \(4x^2-y=100\) with respect to \(t\).
\[\frac{d}{dt}(4x^2-y) = \frac{d}{dt}100\]
\[\frac{d}{dt}(4x^2) - \frac{dy}{dt} = 0\]
Apply chain rule to the first term:
\[\frac{d}{dx}(4x^2) \frac{dx}{dt} - \frac{dy}{dt} = 0\]
Thus,
\[\frac{dy}{dt} = \frac{d}{dx}(4x^2) \frac{dx}{dt}\]

Can you find this now? :)

Answer 4

so you find the derivative of 4x^2 and plug 15 into x then multiple that answer by 8?

Answer 5

Yes.

Answer 6

the answer should be 960