Math Problem Statement

find the derivative of y with respect to the appropriate variable, y = arcsin(√2 t)

Solution

The given function is:

y=arcsin(2t)y = \arcsin(\sqrt{2} t)

We need to find the derivative of yy with respect to tt. Let's differentiate step by step.

Step 1: Differentiate y=arcsin(u)y = \arcsin(u)

The derivative of arcsin(u)\arcsin(u) with respect to uu is:

dduarcsin(u)=11u2\frac{d}{du} \arcsin(u) = \frac{1}{\sqrt{1 - u^2}}

Step 2: Use the Chain Rule

Here, u=2tu = \sqrt{2} t. By the chain rule:

dydt=dduarcsin(u)dudt\frac{dy}{dt} = \frac{d}{du} \arcsin(u) \cdot \frac{du}{dt}

First, let's find dudt\frac{du}{dt}:

dudt=ddt(2t)=2\frac{du}{dt} = \frac{d}{dt} (\sqrt{2} t) = \sqrt{2}

Now substitute into the chain rule:

dydt=11(2t)22\frac{dy}{dt} = \frac{1}{\sqrt{1 - (\sqrt{2} t)^2}} \cdot \sqrt{2}

Step 3: Simplify the Expression

Simplify the term inside the square root:

(2t)2=2t2(\sqrt{2} t)^2 = 2t^2

Thus, the derivative becomes:

dydt=212t2\frac{dy}{dt} = \frac{\sqrt{2}}{\sqrt{1 - 2t^2}}

Final Answer:

The derivative of yy with respect to tt is:

dydt=212t2\frac{dy}{dt} = \frac{\sqrt{2}}{\sqrt{1 - 2t^2}}

Would you like further details or have any questions?


Here are 5 related questions to further explore this topic:

  1. How would you find the second derivative of this function?
  2. What are the conditions for the domain of y=arcsin(2t)y = \arcsin(\sqrt{2} t)?
  3. How does this derivative behave as tt approaches certain critical values (e.g., t=12t = \frac{1}{\sqrt{2}})?
  4. Can we derive this result using implicit differentiation?
  5. How would the derivative change if we had y=arcsin(kt)y = \arcsin(kt) for some constant kk?

Tip:

When differentiating inverse trigonometric functions, it's important to remember to apply the chain rule whenever the argument is not a simple variable.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differentiation
Inverse Trigonometric Functions
Chain Rule

Formulas

d/dx [arcsin(u)] = 1 / √(1 - u^2)
Chain Rule: dy/dx = dy/du * du/dx

Theorems

Derivative of Inverse Sine Function
Chain Rule in Differentiation

Suitable Grade Level

Grades 11-12 or University Calculus