Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Let xx and yy be functions of tt with y=x3+2x+1y = x^3 + 2x + 1. If dxdt=1\frac{dx}{dt} = -1, what is dydt\frac{dy}{dt} when x=4x = 4?\newlineWrite an exact, simplified answer.

Full solution

Q. Let xx and yy be functions of tt with y=x3+2x+1y = x^3 + 2x + 1. If dxdt=1\frac{dx}{dt} = -1, what is dydt\frac{dy}{dt} when x=4x = 4?\newlineWrite an exact, simplified answer.
  1. Apply Chain Rule: To find dydt\frac{dy}{dt}, use the chain rule on y=x3+2x+1y = x^3 + 2x + 1. Differentiate yy with respect to xx to get dydx\frac{dy}{dx}.\newlinedydx=3x2+2\frac{dy}{dx} = 3x^2 + 2.
  2. Substitute x=4x = 4: Substitute x=4x = 4 into dydx\frac{dy}{dx} to find the value at x=4x = 4.dydx=3(4)2+2=3×16+2=48+2=50\frac{dy}{dx} = 3(4)^2 + 2 = 3\times16 + 2 = 48 + 2 = 50.
  3. Use Chain Rule with dxdt\frac{dx}{dt}: Now, use the given dxdt=1\frac{dx}{dt} = -1 to find dydt\frac{dy}{dt} using the chain rule: dydt=dydxdxdt\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}.dydt=50(1)=50\frac{dy}{dt} = 50 \cdot (-1) = -50.

More problems from Complex conjugate theorem