Problem 48 - Calculus and its applications

Question

Find \(\frac{d y}{d u}, \frac{d u}{d x},\) and \(\frac{d y}{d x}\) $$ y=\frac{u+1}{u-1} \text { and } u=1+\sqrt{x} $$

Step-by-Step Solution

Answer

\(\frac{d y}{d u} = \frac{-2}{(u-1)^2}, \frac{d u}{d x} = \frac{1}{2\sqrt{x}}, \frac{d y}{d x} = \frac{-1}{x^{3/2}}\)

Step 1: Differentiate y with respect to u

We start by differentiating the expression for \(y\) with respect to \(u\). Given \( y = \frac{u+1}{u-1} \), let's use the quotient rule for derivatives which states \( \frac{d}{du}\left(\frac{v}{w}\right) = \frac{v'w - vw'}{w^2} \). Here, \( v = u+1 \) and \( w = u-1 \).Quotient rule application:- \( v' = 1, \quad w' = 1 \)Therefore, \[\frac{d y}{d u} = \frac{(1)(u-1) - (u+1)(1)}{(u-1)^2} = \frac{u-1-u-1}{(u-1)^2} = \frac{-2}{(u-1)^2}\]

Step 2: Differentiate u with respect to x

Next, differentiate \(u\) with respect to \(x\) from the relation \( u = 1 + \sqrt{x} \).The derivative of \( \sqrt{x} \) with respect to \(x\) is \( \frac{1}{2\sqrt{x}} \). Thus,\[\frac{d u}{d x} = \frac{d}{dx}(1 + \sqrt{x}) = 0 + \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}}\]

Step 3: Apply the chain rule to find dy/dx

To find \(\frac{d y}{d x}\), use the chain rule, which states: \( \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} \).Substituting the values derived in previous steps:\[\frac{d y}{d x} = \frac{-2}{(u-1)^2} \cdot \frac{1}{2\sqrt{x}}\]Simplifying,\[\frac{d y}{d x} = \frac{-2}{2(u-1)^2\sqrt{x}} = \frac{-1}{(u-1)^2\sqrt{x}}\]

Step 4: Express in terms of x

Since \(u = 1 + \sqrt{x}\), replace \(u\) in the equation for \(\frac{d y}{d x}\) to express it in terms of \(x\).We have: \( u - 1 = \sqrt{x} \).Thus,\[\frac{d y}{d x} = \frac{-1}{(\sqrt{x})^2\sqrt{x}} = \frac{-1}{x\sqrt{x}} = \frac{-1}{x^{3/2}}\]

Key Concepts

Understanding the Chain Rule

In calculus, the chain rule is an essential tool for differentiating composite functions. A composite function, in simple terms, is a function that contains another function within it. The chain rule helps us find derivatives of these nested functions by linking their individual rates of change.
To apply the chain rule, you follow the formula:

• Given a function of the form \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is found by \( \frac{dy}{dg} \cdot \frac{dg}{dx} \).

This means you first take the derivative of the outer function with respect to the inner function. Then multiply it by the derivative of the inner function with respect to its variable, which results in the overall derivative of the function. In our exercise, after identifying \(\frac{dy}{du}\) and \(\frac{du}{dx}\), applying the chain rule seamlessly brings us to the solution for \(\frac{dy}{dx}\).
This method relies on understanding how changes in one variable affect another, everything is connected in a cascade of derivatives.

Mastering the Quotient Rule

The quotient rule is critical when dealing with derivatives of ratios or fractions of functions. When you have a function expressed as a division, like \( y = \frac{v}{w} \), the quotient rule provides us with a technique to find the derivative. The rule states:

• \( \frac{d}{dx}\left(\frac{v}{w}\right) = \frac{v'w - vw'}{w^2} \)

This formula might seem daunting, but it is manageable if broken down. Here, \(v\) is the numerator and \(w\) is the denominator. By calculating the derivatives of both \(v\) and \(w\), which are \(v'\) and \(w'\) respectively, we substitute these into the formula to find the overall derivative.
In our problem, applying this rule allowed us to find \( \frac{dy}{du} \), which was crucial for further applications of the chain rule. Although it may require more steps, the quotient rule ensures accuracy when differentiating any function that involves division.

Derivative of Square Roots Made Easy

Square roots are common in calculus, and understanding their derivatives is important. To find the derivative of \(\sqrt{x}\), we convert it to its equivalent exponent form, \(x^{1/2}\). This transformation simplifies the differentiation process.
Here's how you differentiate \(\sqrt{x}\):

• Apply the power rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \)

By setting \(n = \frac{1}{2}\):
We find \( \frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2}\sqrt{x} \)
This derivative was used in our exercise when finding \(\frac{du}{dx}\), after identifying \(u = 1 + \sqrt{x}\). Understanding these simple rules and transformations allows for quick calculation of derivatives involving square roots.

The Concept of Implicit Differentiation

Implicit differentiation is used when dealing with equations not easily solved for one variable in terms of another. It allows us to differentiate both sides of an equation simultaneously, even when they are interdependent.
In cases where a function is not expressed explicitly as \(y = f(x)\), we can still find \(\frac{dy}{dx}\) indirectly. Here's a simple approach:

• Differentiating each side of the equation, keeping track of each variable's derivative.
• Solve for the derivative of the variable of interest.

Implicit differentiation is powerful when equations involve products or power of terms between both variables, yet its application wasn't directly necessary in our exercise. However, understanding it enriches calculus knowledge, ensuring you are equipped to handle more complex situations where explicit solutions are not easily obtainable.