Problem 24 - Calculus an Applied Approach

Question

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

Step-by-Step Solution

Answer

The derivative is undefined at point (-1,1).

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

Differentiating \((x+y)^{3}\) with respect to x, apply the Chain Rule, yields \(3(x+y)^{2} \cdot (1 + dy/dx)\). Differentiating \(x^{3}+y^{3}\) gives \(3x^{2} + 3y^{2} \cdot dy/dx\). So our equation is now \(3(x+y)^{2} \cdot (1 + dy/dx) = 3x^{2} + 3y^{2} \cdot dy/dx\).

Step 2: Solve for dy/dx

First, we will isolate terms involving \(dy/dx\) on one side of the equation, and everything else on the other side. This gives us \(3(x+y)^{2} \cdot dy/dx - 3y^{2} \cdot dy/dx = 3x^{2} -3(x+y)^{2}\). Factoring out the common factor of dy/dx gives us \(dy/dx = (3x^{2} -3(x+y)^{2}) / (3(x+y)^{2} -3y^{2})\).

Step 3: Evaluate the derivative at the given point

We're given the point (-1,1). We put these values of x and y into the expression for \(dy/dx\) from Step 2, and simplify. This gives us \( dy/dx = [3(-1)^{2} -3((-1+1)^{2})] / [3(-1+1)^{2} -3(1^{2})] = 3/0\).

Key Concepts

Calculus

Calculus is a branch of mathematics that focuses on rates of change and the accumulation of quantities. It is divided into two main parts: differential calculus and integral calculus. Differential calculus deals with the concept of a derivative, which represents an instantaneous rate of change of a function with respect to one of its variables. Implicit differentiation is a technique used in differential calculus to find the derivative of a function that is not explicitly solved for one variable in terms of another. It is particularly useful for equations where the relationship between variables is not straightforwardly given.

Chain Rule

The chain rule is a fundamental theorem in calculus used to compute the derivative of a composition of two or more functions. In basic terms, if we have a function that is composed of other functions, the chain rule helps us differentiate it by taking the derivative of the outer function multiplied by the derivative of the inner function. This is often represented as \( \frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx} \) if \( y \) is a function of \( x \) and \( z \) is a function of \( y \). When working with implicit differentiation, the chain rule allows us to differentiate terms that involve both variables \( x \) and \( y \) by treating \( y \) as a function of \( x \) even if the precise functional relationship is not explicitly known.

Solve for dy/dx

To solve for \( \frac{dy}{dx} \) means to find the derivative of \( y \) with respect to \( x \)—that is, how quickly \( y \) changes as \( x \) changes. In the context of implicit differentiation, after finding the derivative using the chain rule and other differentiation techniques, we need to express \( \frac{dy}{dx} \) explicitly. This means we have to manipulate the equation to isolate \( \frac{dy}{dx} \) on one side. The result gives us a formula that can be used to compute the slope of a tangent to a curve at any given point, provided we know the coordinates of the point.

Evaluate Derivative

To evaluate a derivative at a specific point means to compute the slope of the tangent line to the function at that particular point. When doing implicit differentiation, after finding the expression for \( \frac{dy}{dx} \) in terms of \( x \) and \( y \) variables, we then substitute the coordinates of the given point into this expression. Simplifying the result will yield the numerical value of the slope at that point. However, if the calculation results in an undefined expression such as a division by zero, it means the tangent line at that point is vertical or the function has a cusp or discontinuity at that point.