Problem 35 - Calculus Early Transcendentals

Question

Given the following equations, evaluate \(d y / d x .\) Assume that each equation implicitly defines \(y\) as a differentiable function of \(x\). $$\sqrt{x^{2}+2 x y+y^{4}}=3$$

Step-by-Step Solution

Answer

Question: Find the derivative of \(y\) with respect to \(x\) (\(dy/dx\)) for the given equation: $$\sqrt{x^{2}+2 x y+y^{4}}=3$$ Answer: The derivative of \(y\) with respect to \(x\) is: $$\frac{dy}{dx} =\frac{-2x}{2x + 4y^{3}}$$

Step 1: Differentiate both sides of the equation with respect to x

To find \(dy/dx\), start by differentiating both sides of the given equation with respect to \(x\). Apply the chain rule where necessary: $$\frac{d}{dx}[\sqrt{x^{2}+2 x y+y^{4}}] = \frac{d}{dx}[3]$$

Step 2: Apply the chain rule to the left side of the equation

Use the chain rule to differentiate the left side of the equation. First, differentiate the square root function then multiply by the derivative of the inside function (differentiating the inside function using the product rule and remembering that \(y\) is a function of \(x\)): $$\frac{1}{2}(x^{2}+2 x y+y^{4})^{-\frac{1}{2}} \cdot \frac{d}{dx}[x^{2}+2 x y+y^{4}]= 0$$

Step 3: Differentiate inside of the square root

Differentiate the terms inside the square root using the power and product rule, remembering that \(y\) is a function of \(x\). To differentiate \(2xy\), use the product rule. $$\frac{1}{2}(x^{2}+2 x y+y^{4})^{-\frac{1}{2}} \cdot (2x + 2x\frac{dy}{dx} + 4y^{3}\frac{dy}{dx}) = 0$$

Step 4: Isolate the term containing \(dy/dx\)

In order to find the derivative, isolate the term that contains \(dy/dx\). Start by multiplying both sides by \(2(x^{2}+2 x y+y^{4})^{\frac{1}{2}}\) to get rid of the negative exponent. $$2x + 2x\frac{dy}{dx} + 4y^{3}\frac{dy}{dx} = 0$$ Now, collect the terms containing \(dy/dx\) on one side and the constant term on the other side: $$2x\frac{dy}{dx} + 4y^{3}\frac{dy}{dx} = -2x$$

Step 5: Solve for \(dy/dx\)

Factor out the \(dy/dx\) term, then divide to solve for the expression of the derivative: $$\frac{dy}{dx}(2x + 4y^{3}) = -2x$$ Finally, divide both sides by \((2x + 4y^{3})\) to obtain the expression for \(dy/dx\): $$\frac{dy}{dx} =\frac{-2x}{2x + 4y^{3}}$$

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. When a function is composed of other functions, the chain rule allows us to find its derivative by multiplying the derivatives of these inner functions. It's particularly useful in situations where variables are nested within each other.

To apply the chain rule, you take the derivative of the outer function and then multiply it by the derivative of the inner function. In the context of our exercise, \[ \sqrt{x^{2}+2x y + y^{4}} = 3 \,\] the outer function is the square root, and the inner function is the expression inside the square root \((x^2 + 2xy + y^4)\).

• First, differentiate the outer function, the square root, which gives \[(1/2)(x^{2}+2xy+y^{4})^{-1/2}\].
• Next, multiply by the derivative of the inner function, which involves using the product rule for the \(2xy\) term.

This step is crucial as it links the change in \(y\) with respect to \(x\), hence allowing us to solve for \(dy/dx\).

Product Rule

The product rule is applied when taking the derivative of a product of two functions. It is essential for accurately differentiating terms that multiply, such as \(2xy\), where both \(x\) and \(y\) vary with respect to \(x\).

The rule states that if you have a product of two functions, \(u(x) \,\text{and}\, v(x)\), the derivative is given by:

• Derive the first function and multiply it by the second function: \(u'(x) v(x)\)
• Then, add the product of the derivative of the second function times the first function: \(u(x) v'(x)\)
• Altogether: \[ (uv)' = u'v + uv' \]

In our equation, \(2xy\), we can apply the product rule:

• Differentiate \(x\) as \(1\) and keep \(y\) as it is, getting \(2y\). This represents the change in \(x\).
• Then keep \(x\) and differentiate \(y\) with respect to \(x\), yielding \(2x \frac{dy}{dx}\).

Incorporating the product rule helps in accurately incorporating all variable changes, making it possible to solve for implicit derivatives.

Derivative of Implicit Functions

Implicit differentiation helps find derivatives when \(y\) is not outright isolated. Functions defined implicitly require us to differentiate each term with respect to \(x\) while treating \(y\) as an implicit function of \(x\).

Implicitly defined functions, like our equation \(\sqrt{x^{2}+2xy+y^{4}}=3\), require steps to find \(dy/dx\):

• Differentiating terms directly related to \(x\) uses standard rules, and terms involving \(y\) require a multiplication by \(\frac{dy}{dx}\), indicating that \(y\) is also a variable dependent on \(x\).
• As shown in the exercise, isolate terms involving \(dy/dx\) from those that don't.
• Solve the stepwise differentiated equation for \(dy/dx\), simplifying to \[ \frac{dy}{dx} =\frac{-2x}{2x + 4y^{3}} \]

This process is invaluable for equations where \(y\) cannot be cleanly separated from \(x\), allowing us to still find their rate of change relative to each other.