Problem 70 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=\sqrt{3 t+\sqrt{2+\sqrt{1-t}}}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = \frac{3 - \frac{1}{4\sqrt{2 + \sqrt{1-t}} \cdot \sqrt{1-t}}}{2\sqrt{3t + \sqrt{2 + \sqrt{1-t}}}} \).

Step 1: Recognize the Nested Composition

The function given is a composite function: \( y = \sqrt{3t + \sqrt{2 + \sqrt{1 - t}}} \). We notice it involves nested square roots, which indicates using the chain rule multiple times will be the approach.

Step 2: Differentiate the Outer Function

First, apply the chain rule to the outside function, \( \sqrt{3t + \sqrt{2 + \sqrt{1-t}}} \): \( \frac{d}{dt}\left( \sqrt{3t + \sqrt{2 + \sqrt{1-t}}} \right) = \frac{1}{2\sqrt{3t + \sqrt{2 + \sqrt{1-t}}}} \cdot \frac{d}{dt}(3t + \sqrt{2 + \sqrt{1-t}}) \).

Step 3: Differentiate the Middle Function

Next, differentiate the inside argument of the first square root, \(3t + \sqrt{2 + \sqrt{1-t}}\): For \(3t\), the derivative is \(3\).Now apply the chain rule to \( \sqrt{2 + \sqrt{1-t}} \): \( \frac{d}{dt}\left( \sqrt{2 + \sqrt{1-t}} \right) = \frac{1}{2\sqrt{2 + \sqrt{1-t}}} \cdot \frac{d}{dt}(2 + \sqrt{1-t}) \).

Step 4: Differentiate the Innermost Function

For the \( \sqrt{1-t} \) term, apply the chain rule again:\( \frac{d}{dt}(2 + \sqrt{1-t}) = \frac{1}{2\sqrt{1-t}} \cdot \frac{d}{dt}(1-t) \), where \( \frac{d}{dt}(1-t) = -1 \).Thus, \( \frac{d}{dt}(2 + \sqrt{1-t}) = \frac{-1}{2\sqrt{1-t}} \).

Step 5: Combine Derivatives and Simplify

Substitute back to complete the derivative sequence:From Step 3: \( \frac{d}{dt}(3t + \sqrt{2 + \sqrt{1-t}}) = 3 + \frac{1}{2\sqrt{2 + \sqrt{1-t}}} \cdot \frac{-1}{2\sqrt{1-t}} \).Now combine:\( \frac{d}{dt}(3t + \sqrt{2 + \sqrt{1-t}}) = 3 - \frac{1}{4\sqrt{2+\sqrt{1-t}} \cdot \sqrt{1-t}} \).Substitute this into Step 2 results:\( \frac{dy}{dt} = \frac{1}{2\sqrt{3t + \sqrt{2 + \sqrt{1-t}}}} \cdot \left[3 - \frac{1}{4\sqrt{2+\sqrt{1-t}} \cdot \sqrt{1-t}}\right] \).

Key Concepts

Differentiation

Differentiation refers to the process of finding the derivative of a function. The derivative measures how a function's output changes as its input changes, essentially describing the function's rate of change at any given point. This is a crucial concept in calculus and is widely used to solve various real-world problems, from physics to economics.

In our original exercise, differentiation is applied using the chain rule multiple times because of the nested functions in the equation. The goal is to find an expression for \( \frac{dy}{dt} \), which tells us how the function \( y \) changes with respect to time \( t \).

When performing differentiation, follow these steps:

• Identify the outermost function to differentiate first.
• Apply the chain rule to differentiate the inner functions successively.
• Simplify the result to obtain the final derivative expression.

Differentiation allows us to explore and understand the nature of functions through their instantaneous rate of change.

Composite Function

A composite function is formed when one function is applied to the result of another function. In simpler terms, it's a function inside another function, like layers of an onion. In mathematical notation, if you have two functions, \( f(x) \) and \( g(x) \), a composite function is written as \( f(g(x)) \).

In the given exercise, the function \( y = \sqrt{3t + \sqrt{2 + \sqrt{1 - t}}} \) is a perfect example of a composite function due to its layers of nested square root expressions. Each square root adds a new layer that must be differentiated by peeling back each layer separately using the chain rule.

When dealing with composite functions, remember:

• Each layer or "nest" must be differentiated individually, starting from the outermost layer.
• Use the chain rule, which helps handle these nested functions systematically.

Composite functions allow us to break down complex problems into manageable parts, making it easier to find derivatives and understand how changes in one variable impact the entire function.

Nested Functions

Nested functions occur when a function is placed within another function, leading to a multiple layers or "nests" structure. These require applying the chain rule iteratively for differentiation.

For instance, in the exercise, we have \( y = \sqrt{3t + \sqrt{2 + \sqrt{1 - t}}} \). Here, each square root represents another layer or nest, and we have to apply the chain rule to peel away each one systematically. First, handle the outermost square root, then proceed inward to the next level of nesting, and so on.

When working with nested functions:

• Identify all layers of nesting to determine the number of times you'll apply the chain rule.
• Start differentiating from the outermost nest and work your way to the innermost function.
• Simplify after differentiating to combine the derivatives into a coherent expression.

Nested functions often appear in complex real-world scenarios where a change in one variable affects multiple interrelated processes.