Question

Find \(d y / d x\) by implicit differentiation and evaluate the derivative at the given point. \((x+y)^{3}=x^{3}+y^{3}, \quad(-1,1)\)

Short answer

The derivative of the given function at the point (-1,1) is 0.

Step-by-step solution

Apply implicit differentiation to both sides

Differentiate both sides of the equation \((x+y)^{3} = x^{3}+y^{3}\). Use the chain rule when doing so, i.e. let \(u = x+y\) then differentiate \(u^{3}\). For RHS, each term can be differentiated separately. Thus, we get `3(x+y)^2 * dy/dx = 3x^2 + 3y^2 * dy/dx`

Solve for \( dy/dx \)

Rearrange terms to isolate \( dy/dx \) on one side of the equation by subtracting `3x^2` from both sides and factoring out `dy/dx`. This results in `dy/dx = (3x^2 - 3(x+y)^2) / (3(x+y)^2 - 3y^2)`.

Simplify \( dy/dx \) equation

Simplify the equation of \(dy/dx\) further by cancelling out common values and combine like terms. The final equation will be `dy/dx = (x^2 - (x+y)^2) / ((x+y)^2 - y^2)`.

Evaluate \( dy/dx \) at the given point

Lastly, substitute the given values of x = -1 and y = 1 into the equation of \(dy/dx\). The \(dy/dx\) at point (-1,1) results in `0`.

Key concepts

Understanding the Chain Rule in Implicit Differentiation

When differentiating equations that involve several variables and functions of those variables, the chain rule is essential. It helps us find the derivative of composite functions where one function is nested inside another.

In the given problem, the function \((x+y)^3\) suggests that we are dealing with a composite function. Here, we treat \(x+y\) as a single unit or variable, say \(u\), making the function \(u^3\). The chain rule allows us to differentiate \(u^3\) by taking the derivative of the outer function and then multiplying it by the derivative of the inner function: \[\frac{d}{du}(u^3) = 3u^2 \cdot \frac{du}{dx}.\]

This step is crucial in implicit differentiation because, in implicit equations, the variables are not separated, and differentiating traditionally becomes challenging. By utilizing the chain rule, we handle such combinations effectively.

Simplifying Derivative Evaluation with Implicit Differentiation

Derivative evaluation involves calculating the rate of change of a function concerning a specific variable. In implicit differentiation, it's all about finding how one variable changes concerning another, even when the function is not solved explicitly for one variable.

In the problem, we aim to find \(\frac{dy}{dx}\) by differentiating both sides of the equation \((x+y)^3 = x^3 + y^3.\) We perform these operations:

• Differentiating the left side using the chain rule produces \(3(x+y)^2 \cdot \frac{dy}{dx}.\)
• The right side involves standard differentiation rules for each term, giving us \(3x^2 + 3y^2 \cdot \frac{dy}{dx}.\)

Our main challenge is to isolate \(\frac{dy}{dx}\). This requires rearranging and factoring, leading to the simplified expression derived in the solution. Evaluating this at a particular point, like \((-1,1),\) provides a numerical value for the derivative. This process captures how one variable affects or relates to another in a specific context.

Implicit Function Theorem: The Backbone of Implicit Differentiation

The implicit function theorem provides the theoretical foundation for derivations in implicit differentiation. This powerful tool helps us understand the conditions under which a relationship defined by an equation can implicitly define a function of a variable.

In simpler terms, even if we can't express \(y\) explicitly as \(f(x),\) the implicit function theorem guarantees the existence of derivatives like \(\frac{dy}{dx}.\) In our exercise, this theorem ensures that we can treat \(y\) as a function of \(x\) and differentiate accordingly.

By using this theorem, and by following steps such as applying the chain rule and rearranging the results, we can confidently find \(\frac{dy}{dx}\) at desired points. This backbone principle allows for a robust approach to dealing with relationships in their naturally complex implicit forms, all while maintaining mathematical rigor.