Computer Explorations


Use a CAS to perform the following steps in Exercises 55–62.


b. Using implicit differentiation, find a formula for the derivative dy/dx and evaluate it at the given point P.


2y² + (xy)¹/³ = x² + 2, P(1,1)

Verified step by step guidance

1. Start by differentiating both sides of the equation 2y² + (xy)^(1/3) = x² + 2 with respect to x. Remember that y is a function of x, so you'll need to use implicit differentiation.
2. Differentiate the left side: For 2y², use the chain rule to get 4y(dy/dx). For (xy)^(1/3), use the chain rule and product rule: differentiate (xy)^(1/3) with respect to x, which involves differentiating the inside function xy.
3. Differentiate the right side: The derivative of x² is 2x, and the derivative of the constant 2 is 0.
4. Set the derivatives equal: Combine the results from the left and right sides to form the equation 4y(dy/dx) + (1/3)(xy)^(-2/3)(y + x(dy/dx)) = 2x.
5. Solve for dy/dx: Isolate dy/dx on one side of the equation. Substitute the point P(1,1) into the equation to evaluate the derivative at that point.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. It involves differentiating both sides of an equation with respect to a variable, typically x, while treating other variables as implicit functions of x. This method is essential when dealing with equations where y cannot be easily isolated.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This rule is crucial when applying implicit differentiation, especially when differentiating terms like (xy)^(1/3), where both x and y are functions of another variable.

Evaluating Derivatives at a Point

Once the derivative dy/dx is found using implicit differentiation, it can be evaluated at a specific point to find the slope of the tangent line at that point. This involves substituting the coordinates of the given point, such as P(1,1), into the derivative formula. This step is important for understanding the behavior of the function at specific locations on its graph.