Problem 3 - Calculus Single Variable

Question

\( y\) is a function of \(x .\) Calculate the derivative of the given expression with respect to \(x\). (Your answer should contain the term \(d y / d x .)\) \(2 x / \sqrt{y}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{2\sqrt{y} - \frac{x}{\sqrt{y}} \cdot \frac{dy}{dx}}{y} \).

Step 1: Identify the Derivative Rule

We need to find the derivative of the expression \( \frac{2x}{\sqrt{y}} \) with respect to \( x \). This expression is a quotient, so we'll use the quotient rule: \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \) where \( u = 2x \) and \( v = \sqrt{y} \).

Step 2: Differentiate the Numerator

Differentiate the numerator \( u = 2x \) with respect to \( x \). The derivative is straightforward: \( \frac{du}{dx} = 2 \).

Step 3: Differentiate the Denominator

Differentiate the denominator \( v = \sqrt{y} = y^{1/2} \) with respect to \( x \). Use the chain rule: \( \frac{d}{dx}(y^{1/2}) = \frac{1}{2}y^{-1/2} \cdot \frac{dy}{dx} \). So, \( \frac{dv}{dx} = \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \).

Step 4: Apply the Quotient Rule

Substitute the derivatives into the quotient rule: \( \frac{d}{dx}\left( \frac{2x}{\sqrt{y}} \right) = \frac{\sqrt{y}(2) - 2x\left( \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \right)}{y} \).

Step 5: Simplify the Expression

Simplify the derivative expression: \( \frac{d}{dx}\left( \frac{2x}{\sqrt{y}} \right) = \frac{2\sqrt{y} - \frac{x}{\sqrt{y}} \cdot \frac{dy}{dx}}{y} \). Factor out terms as needed and adjust for a clean expression.

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental tool for finding the derivative of a quotient of two functions. It’s especially useful when you have one function divided by another, like in the expression \( \frac{2x}{\sqrt{y}} \).

When we have a function \( \frac{u}{v} \), where \( u \) and \( v \) are both differentiable functions of \( x \), the Quotient Rule formula for the derivative is:

• \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \)

For instance, with \( u = 2x \) and \( v = \sqrt{y} \), the rule helps us balance the impact of changing both the numerator and the denominator as \( x \) changes.

By applying the Quotient Rule, we ensure that both parts' derivatives are appropriately accounted for, maintaining the connection between the functions. The derivative \( \frac{du}{dx} = 2 \) and \( \frac{dv}{dx} \) (which requires the Chain Rule) then get plugged into the formula, helping us find the correct derivative of the whole expression.

Chain Rule

The Chain Rule is indispensable when dealing with nested functions—functions within other functions. This rule allows us to differentiate complex compositions by breaking them into simpler parts.

Let's look into the expression \( \sqrt{y} \), which is reformulated as \( y^{1/2} \). This is where the Chain Rule shines:

• For a function \( f(g(x)) \), the derivative \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \).

In our given problem, \( v = y^{1/2} \), makes use of the Chain Rule to consider \( y \) as an internal component of \( y^{1/2} \). Thus, by applying the Chain Rule, \( \frac{d}{dx}(y^{1/2}) = \frac{1}{2}y^{-1/2} \cdot \frac{dy}{dx} \) allows us to neatly incorporate the rate of change of \( y \) with respect to \( x \).

Understanding and deploying the Chain Rule correctly is crucial here, as it captures the way one variable affects another through a more complex relationship. It reminds us to multiply by the derivative of the inner function, \( \frac{dy}{dx} \), highlighting the indirect impact of \( x \) on \( y \). This makes our derivative complete and correct within the context of implicit functions.

Implicit Differentiation

Implicit Differentiation is a technique used when functions are intertwined rather than isolated as \( y = f(x) \). It's particularly necessary when dealing with equations that define \( y \) implicitly rather than explicitly.

In the given problem, \( y \) is a function of \( x \), and it’s not directly solved for \( y \) beforehand. Implicit differentiation allows us to differentiate with respect to \( x \) while treating \( y \) as an implicit function that also depends on \( x \).

Here's how it plays into the solution:

• Whenever you differentiate \( y \) with respect to \( x \), attach a \( \frac{dy}{dx} \) tag to denote this relationship.

This approach helps calculate derivatives of terms like \( y^{1/2} \) through using implicit differentiation in tandem with the Chain Rule. As a result, it makes sure that the hidden dependence of \( y \) on \( x \) is properly represented with terms like \( \frac{dy}{dx} \).

Implicit Differentiation ensures we don’t lose track of variables' relationships even when they are not isolated. It's a critical mechanism in examining how multi-variable equations dynamically adjust as one variable changes, particularly valuable in finding derivatives for non-explicit functions.