Problem 305 - Calculus Volume 1

Question

For the following exercises, use implicit differentiation to find \(\frac{d y}{d x}\) $$ y \sqrt{x+4}=x y+8 $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{y (1 - \frac{1}{2\sqrt{x+4}})}{\sqrt{x+4} - x} \)

Step 1: Differentiate Both Sides

Differentiate both sides of the equation with respect to \( x \). Start with the left-hand side. Use the product rule for differentiation on \( y \sqrt{x+4} \): \[\frac{d}{dx}\left(y \sqrt{x+4}\right) = y \cdot \frac{d}{dx}\left(\sqrt{x+4}\right) + \sqrt{x+4} \cdot \frac{dy}{dx}\]For the right-hand side, differentiate \( xy + 8 \) using the product rule and constant rule: \[\frac{d}{dx}\left(xy + 8\right) = y + x \cdot \frac{dy}{dx}\]

Step 2: Compute Specific Derivatives

Find the derivative of \( \sqrt{x+4} \) using the chain rule: \[\frac{d}{dx}\left(\sqrt{x+4}\right) = \frac{1}{2\sqrt{x+4}}\]Now substitute this back into the equation for the derivative of the left-hand side:\[\frac{d}{dx}\left(y \sqrt{x+4}\right) = y \cdot \frac{1}{2\sqrt{x+4}} + \sqrt{x+4} \frac{dy}{dx}\]

Step 3: Assemble the Full Derivative Equation

Write down the full derivative equation combining the derivatives from both sides:\[y \cdot \frac{1}{2\sqrt{x+4}} + \sqrt{x+4} \frac{dy}{dx} = y + x \frac{dy}{dx}\]

Step 4: Isolate \(\frac{dy}{dx}\)

Rearrange the terms to solve for \(\frac{dy}{dx}\). Move all terms involving \(\frac{dy}{dx}\) to one side of the equation while keeping others on the opposite side:\[\sqrt{x+4} \frac{dy}{dx} - x \frac{dy}{dx} = y - y \cdot \frac{1}{2\sqrt{x+4}}\]

Step 5: Factor and Solve for \(\frac{dy}{dx}\)

Factor out \(\frac{dy}{dx}\) from the left side:\[\left(\sqrt{x+4} - x\right) \frac{dy}{dx} = y - y \cdot \frac{1}{2\sqrt{x+4}}\]Finally, solve for \(\frac{dy}{dx}\) by dividing both sides:\[\frac{dy}{dx} = \frac{y - \frac{y}{2\sqrt{x+4}}}{\sqrt{x+4} - x}\]

Step 6: Simplify the Expression

Further simplify the numerator by factoring out \(y\):\[\frac{dy}{dx} = \frac{y \left(1 - \frac{1}{2\sqrt{x+4}}\right)}{\sqrt{x+4} - x}\]

Key Concepts

Product Rule

When dealing with the product rule, it's all about differentiating products of two functions. The product rule tells us how to differentiate functions that are multiplied together. Let's say we have two functions, \( u(x) \) and \( v(x) \). The product rule states that: \[\frac{d}{dx}(uv) = u'v + uv'\] This means we differentiate \( u \) while keeping \( v \) steady, and then differentiate \( v \) while keeping \( u \) steady, and sum both results. In our problem, we apply the product rule to \( y \sqrt{x+4} \):

• Differentiating \( y \) times \( \sqrt{x+4} \) involves taking the derivative of \( y \) (giving \( \frac{dy}{dx} \)) while keeping \( \sqrt{x+4} \) constant.
• Then, differentiate \( \sqrt{x+4} \) while keeping \( y \) constant.

This technique allows us to break down complex differentiation into simpler parts, making it easier to calculate the derivative.

Chain Rule

The chain rule is a critical tool in calculus for differentiating composite functions, which are functions within functions. Suppose you have \( f(g(x)) \), a composition of \( f \) and \( g \). The chain rule says: \[\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\] This rule helps us separate and tackle each function individually. In our implicit differentiation exercise, we use the chain rule to differentiate \( \sqrt{x+4} \):

• Recognize \( \sqrt{x+4} \) as a composition of an outer function \( f(u) = \sqrt{u} \) and an inner function \( g(x) = x+4 \).
• The derivative of \( \sqrt{x+4} \) is computed as \( \frac{1}{2\sqrt{x+4}} \), following the chain rule steps: \( f'(u) = \frac{1}{2\sqrt{u}} \) and \( g'(x) = 1 \).

This allows us to handle complex expressions efficiently, especially when paired with the product rule.

Solving Derivatives

Solving derivatives in implicit differentiation involves finding \( \frac{dy}{dx} \) when \( y \) and \( x \) are mingled in an equation. Instead of making \( y \) a function of \( x \), we differentiate both sides of the equation directly with respect to \( x \). The process involves:

• Applying differentiation rules like the product and chain rules to both sides of the given equation.
• After finding the derivative expressions, rearranging the equation to isolate \( \frac{dy}{dx} \).
• Factoring and simplifying the resulting expression to solve for \( \frac{dy}{dx} \).

In our example, \( \frac{dy}{dx} \) was isolated and solved step by step, resulting in: \[\frac{dy}{dx} = \frac{y \left(1 - \frac{1}{2\sqrt{x+4}}\right)}{\sqrt{x+4} - x}\] This expression gives us the derivative of \( y \) with respect to \( x \), capturing how changes in \( x \) affect \( y \). Understanding and applying these techniques accurately is crucial to mastering calculus problems that involve implicit differentiation.