Problem 22 - Thomas Calculus

Question

Use implicit differentiation to find \(d y / d x\) in Exercises \(19-32\) $$ x^{3}-x y+y^{3}=1 $$

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{dx} = \frac{y - 3x^2}{-x + 3y^2}\).

Step 1: Differentiate the Expression with Respect to x

Begin by differentiating both sides of the equation \(x^3 - xy + y^3 = 1\) with respect to \(x\). Remember to treat \(y\) as an implicit function of \(x\). The derivative of \(x^3\) is \(3x^2\).

Step 2: Apply Product Rule to -xy

For the term \(-xy\), apply the product rule: differentiate \(x\) to get \(1\) while keeping \(y\) constant, then differentiate \(y\) to get \(\frac{dy}{dx}\) while keeping \(-x\) as a constant multiplier. This gives \(-y - x\frac{dy}{dx}\).

Step 3: Differentiate y^3 Using Chain Rule

Differentiate \(y^3\) with respect to \(x\) using the chain rule: \(3y^2\frac{dy}{dx}\).

Step 4: Combine Differentiated Terms

Combine all the differentiated parts: \(3x^2 - y - x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0\).

Step 5: Solve for dy/dx

Rearrange the equation to solve for \(\frac{dy}{dx}\). First, gather all \(\frac{dy}{dx}\) terms on one side and move other terms to the opposite side: \(-x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = y - 3x^2\). Factor \(\frac{dy}{dx}\) out: \(\frac{dy}{dx}(-x + 3y^2) = y - 3x^2\).

Step 6: Isolate dy/dx

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

Key Concepts

Product Rule

When faced with a multiplication of two functions, such as the term \(-xy\), you need to use the product rule for differentiation. The product rule states: if you have two functions \(u\) and \(v\), the derivative of their product \(uv\) is \(u'v + uv'\). In this exercise, \(x\) can be viewed as \(u\) and \(y\) as \(v\). First, differentiate the \(x\) term to get 1 while treating \(y\) as a constant. Then differentiate \(y\) to get \(\frac{dy}{dx}\) while considering \(-x\) as a constant multiplier. Therefore, the derivative of \(-xy\) becomes \(-y - x\frac{dy}{dx}\). This method helps to manage different parts of a complex expression easily while maintaining accuracy when differentiating hybrid functions consisting of more than one variable.

Chain Rule

The chain rule is essential when dealing with composite functions. A composite function is when you have a function inside another function, like \(y^3\) in our exercise. According to the chain rule, if you have a composite function \(f(g(x))\), the derivative is \(f'(g(x)) \times g'(x)\). In this context, \(f(u) = u^3\) where \(u = y(x)\). The outer function derivative \(f'(u)\) is \(3u^2\) and the inner function \(u'(x)\) gives \(\frac{dy}{dx}\). Thus, the chain rule gives us \(3y^2 \cdot \frac{dy}{dx}\) for \(y^3\). This method allows the transformation of complex derivatives into simpler forms using a systematic approach.

Derivatives

Derivatives are one of the fundamental tools in calculus used to find the rate at which one quantity changes with respect to another. When differentiating the equation \(x^3 - xy + y^3 = 1\), each term follows individual derivative rules. The derivative of \(x^3\) is straightforward: differentiate it normally to get \(3x^2\). This represents the rate of change of \(x^3\) concerning \(x\). For the term \(-xy\), derivatives become slightly more involved, requiring the product and chain rules to separate and correctly differentiate each part. Collectively, these rules make it feasible to differentiate complex and multi-variable expressions efficiently.

dy/dx Calculation

Finding \(\frac{dy}{dx}\) represents calculating the derivative of \(y\) with respect to \(x\). It indicates how \(y\) changes as \(x\) changes in implicit differentiation contexts. In the provided exercise, we gather terms involving \(\frac{dy}{dx}\) on one side: \(-x\frac{dy}{dx} + 3y^2\frac{dy}{dx}\). Them without \(\frac{dy}{dx}\) on the other side: \(y - 3x^2\). Factor out \(\frac{dy}{dx}\) to simplify: \(\frac{dy}{dx}(-x + 3y^2) = y - 3x^2\). Finally, solve for \(\frac{dy}{dx}\) by dividing both sides by \(-x + 3y^2\), yielding: \(\frac{dy}{dx} = \frac{y - 3x^2}{-x + 3y^2}\). This isolates \(\frac{dy}{dx}\), providing a clear expression for the derivative. This calculation showcases how implicit differentiation enables finding derivatives of functions not explicitly solved for one variable.