Problem 40 - Calculus for Biology and Medicine

Question

In Problems \(40-46\), find \(\frac{d y}{d x}\) by applying the chain rule repeatedly. y=\left(\sqrt{1-2 x^{2}}+1\right)^{2}

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} = -4x - \frac{4x}{\sqrt{1-2x^2}} \).

Step 1: Identify the Composition of Functions

Notice that the given function can be rewritten as a composition of an outer and an inner function. The function is presented as \[ y = ig( u(x) \big)^2 \] where \[ u(x) = \sqrt{1-2x^2} + 1. \] We have a composition where the outer function is \( v(x) = x^2 \) and the inner function is \( u(x) = \sqrt{1-2x^2} + 1 \). We need to apply the chain rule to differentiate it.

Step 2: Differentiate the Outer Function

The outer function is \( v(u) = u^2 \). Differentiating \( v(u) \) with respect to \( u \) gives:\[ \frac{dv}{du} = 2u. \]

Step 3: Differentiate the Inner Function

Next, differentiate the inner function \( u(x) = \sqrt{1-2x^2} + 1 \) with respect to \( x \).First, consider \( g(x) = \sqrt{1-2x^2} \). We apply the chain rule:\[ \frac{d}{dx}(g(x)) = \frac{1}{2\sqrt{1-2x^2}} \cdot \frac{d}{dx}(1-2x^2) = \frac{1}{2\sqrt{1-2x^2}} \cdot (-4x) = \frac{-2x}{\sqrt{1-2x^2}}. \]Since the \( +1 \) is a constant, it has a derivative of 0:\[ \frac{du}{dx} = \frac{-2x}{\sqrt{1-2x^2}}. \]

Step 4: Apply the Chain Rule

Now, apply the chain rule to find \( \frac{dy}{dx} \):\( \frac{dy}{dx} = \frac{dv}{du} \times \frac{du}{dx} = 2u \times \frac{-2x}{\sqrt{1-2x^2}}. \)Substitute back \( u = \sqrt{1-2x^2} + 1 \):\[ \frac{dy}{dx} = 2(\sqrt{1-2x^2} + 1) \cdot \frac{-2x}{\sqrt{1-2x^2}}. \]

Step 5: Simplify the Derivative

Simplify the expression for \( \frac{dy}{dx} \):Expanding the expression:\[ \frac{dy}{dx} = -4x \frac{\sqrt{1-2x^2} + 1}{\sqrt{1-2x^2}}. \]Rewrite and simplify:\[ \frac{dy}{dx} = -4x \frac{(\sqrt{1-2x^2} + 1)}{\sqrt{1-2x^2}}. \]Thus, the derivative is:\[ \frac{dy}{dx} = -4x - \frac{4x}{\sqrt{1-2x^2}}. \]

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used when differentiating a function that is the composition of two or more simpler functions. Imagine you have a function that is nested inside another function. The chain rule helps us find the derivative of these composite functions by working from the outside in.
The basic idea can be understood as follows:

• If you have a composite function \( y = f(g(x)) \),
• and you want to find its derivative \( \frac{dy}{dx} \),
• the chain rule tells you to multiply the derivative of the outer function \( f \) (with respect to its argument) by the derivative of the inner function \( g \) (with respect to \( x \)).

This is expressed in the formula: \[ \frac{dy}{dx} = \frac{dy}{dg} \times \frac{dg}{dx} \]
In more practical terms, it means find the derivative of the outside, times the derivative of the inside, all evaluated at the current "place" (i.e., values of x). This powerful rule allows us to handle complex derivatives easily.

Composition of Functions

A composition of functions involves plugging one function into another. For example, in the function \( y = (\sqrt{1-2x^{2}}+1)^{2} \), we have an outer function and an inner function.
Here's how to break it down:

• The 'inner function' \( u(x) = \sqrt{1-2x^{2}} + 1 \) is evaluated first.
• Then, this result becomes the input for the 'outer function' \( v(u) = u^2 \).

This nested structure is what makes it necessary to apply the chain rule when differentiating.
Understanding this structure is key to effectively using the chain rule. It is important to correctly identify the roles of the outer and inner functions to correctly compose the chain for differentiation. The relationship and dependency between these functions is what makes the chain rule applicable.

Derivative Simplification

Simplifying a derivative involves reducing your final expression to make it as neat and understandable as possible. After applying the chain rule, you may end up with quite a complex expression.
For example, after differentiating the function \( y = (\sqrt{1-2x^2} + 1)^2 \), we get:

• Initial derivative: \[ \frac{dy}{dx} = 2u \cdot \frac{-2x}{\sqrt{1-2x^2}} \]
• Substitute \( u = \sqrt{1-2x^2} + 1 \): \[ \frac{dy}{dx} = -4x \frac{(\sqrt{1-2x^2} + 1)}{\sqrt{1-2x^2}} \]
• Simplified result: \[ \frac{dy}{dx} = -4x - \frac{4x}{\sqrt{1-2x^2}} \]

This simplification helps in clearly seeing how the components of the original functions influence the rate of change of \( y \).
The goal of simplification is to present the derivative in a more concise form, making it easier to understand and work with in future calculations. It often involves factoring, cancelling terms, or rewriting expressions to avoid complex fractions.