Problem 7 - Calculus Early Transcendentals

Question

Find \(d y / d x\) by implicit differentiation. \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=1\)

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \left(\frac{y}{x}\right)^{3/2}\)

Step 1: Differentiate both sides with respect to x

We start by taking the derivative with respect to \(x\) of the equation \(\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = 1\). The derivative of a constant (1) with respect to \(x\) is 0.

Step 2: Differentiate each term on the left

Apply the derivative separately: for \(\frac{1}{\sqrt{x}}\), use the chain rule. Let \(u = \sqrt{x}\), then \(u = x^{1/2}\). The derivative is \(-\frac{1}{2}x^{-3/2}\). For \(\frac{1}{\sqrt{y}}\), use substitution \(v = \sqrt{y}\), then \(v = y^{1/2}\). The derivative is \(-\frac{1}{2}y^{-3/2} \cdot \frac{dy}{dx}\).

Step 3: Write the full equation

Combine the derivatives: \(-\frac{1}{2}x^{-3/2} - \frac{1}{2}y^{-3/2} \frac{dy}{dx} = 0\).

Step 4: Solve for \(\frac{dy}{dx}\)

Rearrange the equation to solve for \(\frac{dy}{dx}\). Add \(\frac{1}{2}y^{-3/2} \frac{dy}{dx}\) to both sides: \(\frac{1}{2}y^{-3/2} \frac{dy}{dx} = \frac{1}{2}x^{-3/2}\).

Step 5: Isolate \(\frac{dy}{dx}\)

Divide both sides by \(\frac{1}{2}y^{-3/2}\) to isolate \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{x^{-3/2}}{y^{-3/2}}\).

Step 6: Simplify the derivative

Simplify the expression: \(\frac{dy}{dx} = \left(\frac{y}{x}\right)^{3/2}\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of composite functions. Imagine a scenario where one function is nested inside another, similar to peeling layers of an onion.
For the exercise given, each part such as \(\frac{1}{\sqrt{x}}\) can be rewritten using exponent notation, \(x^{-1/2}\), making it easier to differentiate using the chain rule.

• Identify the function within a function, like \( \sqrt{x} = x^{1/2} \).
• Find the derivative of the outer function while keeping the inner function the same.
• Multiply by the derivative of the inner function.

In mathematical terms, the chain rule is applied as: If \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\).
Applying this to \(\frac{1}{\sqrt{x}} = x^{-1/2}\), we find \(-\frac{1}{2}x^{-3/2}\), considering both the exponent rule and the chain rule.

Derivative

The concept of a derivative is central to calculus and describes the rate at which a function is changing at any given point.
In simple terms, it tells you what the slope of the tangent line to the function is.

• The derivative of \(x\) with respect to \(x\) is \(1\).
• The derivative of a constant is always \(0\).
• When differentiating, consider whether any function has specific differentiation rules applying to it, like powers or products of functions.

For the given exercise, distinguishing functions like \(x \) and \( y \) in their exponent forms aids in leveraging their respective formulaic derivatives.
Understanding how derivatives behave with respect to different parts of a function, such as \(\frac{1}{2}x^{-3/2}\) simplifies solving equations using derivatives.

Solving for dy/dx

Once you've differentiated each term in an equation, the goal often is to solve for \( \frac{dy}{dx} \), which represents the derivative of \(y\) with respect to \(x\).
This process involves algebraic manipulation of the derived terms.

Steps to Solve for \( \frac{dy}{dx} \)

• Combine like terms and ensure all \( \frac{dy}{dx} \) terms are on one side.
• Isolate the \( \frac{dy}{dx} \) term using algebra, typically involving division or multiplication.
• Simplify the fraction or expression that results from isolating \( \frac{dy}{dx} \).

In our example, we rearrange and manipulate terms so \( \frac{1}{2}y^{-3/2} \cdot \frac{dy}{dx} = \frac{1}{2}x^{-3/2}\).
Then, by isolating \( \frac{dy}{dx} \), it simplifies to \( \frac{dy}{dx} = \left(\frac{y}{x}\right)^{3/2}\).
Through this, the expression \( \frac{dy}{dx}\) gives us how \(y\) changes in response to a small change in \(x\).