Problem 32 - 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 \(x\). $$x^{3}+3 x y^{2}-y^{5}=0$$

Step-by-Step Solution

Answer

Question: Find the derivative of \(y\) with respect to \(x\) for the given equation \(x^3 + 3xy^2 - y^5 = 0\). Answer: \(\frac{dy}{dx} = \frac{3x^2+3y^2}{5y^4-6xy}\)

Step 1: Differentiate both sides with respect to \(x\)

Differentiating both sides of the equation \(x^3 + 3xy^2 - y^5 = 0\) with respect to \(x\) gives us: $$ \frac{d}{dx}(x^3) + \frac{d}{dx}(3xy^2) - \frac{d}{dx}(y^5) = \frac{d}{dx}(0) $$

Step 2: Apply the derivative rules

Use the power rule for the first term, product rule for the second term and chain rule for the third term. We will get: $$ 3x^2 + \left(3y^2 + 6xy\frac{dy}{dx}\right) - 5y^4\frac{dy}{dx} = 0 $$

Step 3: Solve for \(dy/dx\)

Now, we need to isolate the \(dy/dx\) term. We can rewrite the previous equation as follows: $$ 3x^2 + 3y^2 + 6xy\frac{dy}{dx} - 5y^4\frac{dy}{dx} = 0 $$ Next, move all the terms involving \(dy/dx\) to the right side of the equation and factor out \(dy/dx\): $$ 3x^2 + 3y^2 = (5y^4 - 6xy) \frac{dy}{dx} $$ Finally, divide both sides by \((5y^4 - 6xy)\) to obtain the expression for \(dy/dx\): $$ \frac{dy}{dx} = \frac{3x^2+3y^2}{5y^4-6xy} $$ Now, we have found the derivative of \(y\) with respect to \(x\).

Key Concepts

Derivative Rules

When working with implicit differentiation, it's crucial to apply various derivative rules to successfully find derivatives. Derivative rules help us systematically differentiate expressions and perform calculations accurately.

Some of the essential rules include:

• Power Rule: Used for powers of a variable, this rule states that if you have an expression of the form \(x^n\), the derivative is \(nx^{n-1}\).
• Product Rule: This rule is necessary when differentiating the product of two functions, \(u(x)\) and \(v(x)\). The product rule formula is \( (uv)' = u'v + uv' \).
• Chain Rule: The chain rule is used to differentiate composite functions. If you have \(f(g(x))\), the chain rule states the derivative is \(f'(g(x))g'(x)\).

Understanding these derivative rules ensures you can handle complex equations, especially in implicit differentiation settings. Each rule plays a unique role in calculating the derivative of a given equation.

Power Rule

The power rule is one of the fundamental rules in differentiation. It's commonly used because many functions involve polynomials or powers of variables.

To apply the power rule, follow these steps:

• Identify terms in the form of \(x^n\).
• In the derivative, multiply the term by the exponent \(n\), then decrement the exponent by one.
• For example, if \(y = x^3\), its derivative \( \frac{d}{dx}(x^3) = 3x^2 \).

In the original exercise, the power rule was applied to differentiate \(x^3\), resulting in \(3x^2\). This straightforward process highlights why the power rule is so valuable in calculus.

Product Rule

The product rule comes into play when you need to differentiate products of two functions. In implicit differentiation, this rule is especially useful for terms where variables multiply each other, like "3xy^2".

For the product rule, remember the formula:

• \((uv)' = u'v + uv'\)

where \(u\) and \(v\) are functions of \(x\). This means you apply the derivative to each function, keeping the other constant.

In our example, for the term \(3xy^2\), interpret it as a product of \(u = 3x\) and \(v = y^2\) and apply:

• \(u' = 3\), \(v' = 2y\frac{dy}{dx}\)
• The derivative becomes \(3y^2 + 6xy\frac{dy}{dx}\)

Notice how the product rule helps differentiate even with a variable squared in the second term.

Chain Rule

The chain rule is crucial when dealing with compositions of functions, especially in implicit differentiation where terms like \(y^5\) are encountered, requiring careful attention.

When using the chain rule:

• Identify the composition of functions, like \(f(g(x))\).
• Differentiate the outer function first, and then multiply by the derivative of the inner function.
• For example, \((y^5)'\) requires considering \(y\) as a function of \(x\).

For the chain rule in action, see how \( \frac{d}{dx}(y^5) \) was differentiated into \(-5y^4 \frac{dy}{dx}\), capturing both the power and the derivative of \(y\) with respect to \(x\).

This aptitude for decomposing complex expressions into simpler parts makes the chain rule indispensable.