Math Problem Statement

y = √ e^−2t^2 + 4 ,dy/dt =

Solution

To find dydt\frac{dy}{dt}, we need to use the chain rule. The given function is:

y=e2t2+4y = \sqrt{e^{-2t^2} + 4}

Step 1: Apply the chain rule

The derivative of y=uy = \sqrt{u} with respect to tt is:

dydt=12ududt\frac{dy}{dt} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dt}

Here, u=e2t2+4u = e^{-2t^2} + 4. Let's find dudt\frac{du}{dt}.

Step 2: Differentiate uu with respect to tt

u=e2t2+4u = e^{-2t^2} + 4

The derivative of e2t2e^{-2t^2} with respect to tt is found using the chain rule:

[ \frac{d}{dt}[e^{-2t^2}] = e^{-2t^2} \cdot \frac{d}{dt}[-2t^2] = e^{-2t^2} \cdot (-4t) ]

The derivative of the constant 44 is 00. Thus:

dudt=4te2t2\frac{du}{dt} = -4t \cdot e^{-2t^2}

Step 3: Combine results

Now substitute uu and dudt\frac{du}{dt} back into the chain rule formula:

dydt=12e2t2+4(4te2t2)\frac{dy}{dt} = \frac{1}{2\sqrt{e^{-2t^2} + 4}} \cdot (-4t \cdot e^{-2t^2})

Simplify:

dydt=4te2t22e2t2+4\frac{dy}{dt} = \frac{-4t \cdot e^{-2t^2}}{2\sqrt{e^{-2t^2} + 4}}

dydt=2te2t2e2t2+4\frac{dy}{dt} = \frac{-2t \cdot e^{-2t^2}}{\sqrt{e^{-2t^2} + 4}}

Final Answer:

dydt=2te2t2e2t2+4\frac{dy}{dt} = \frac{-2t \cdot e^{-2t^2}}{\sqrt{e^{-2t^2} + 4}}

Let me know if you'd like further clarification or details on any step!


Relative Questions:

  1. How does the chain rule apply in similar cases involving square roots?
  2. What is the derivative of ektne^{kt^n} for any constants kk and nn?
  3. How would the answer change if the square root was replaced by a cube root?
  4. What is the significance of the e2t2e^{-2t^2} term in determining the behavior of dydt\frac{dy}{dt}?
  5. How do we handle higher-order derivatives for this function?

Tip: Always simplify intermediate results before substituting back to avoid unnecessary complexity!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Exponential Functions
Square Root Functions

Formulas

dy/dt = (1 / 2√u) * (du/dt)
d/dt[e^(-kt^n)] = e^(-kt^n) * (-k * n * t^(n-1))

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12