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.


x√(1 + 2y) + y = x², P(1,0)

Verified step by step guidance

1. Start by differentiating both sides of the equation with respect to x. The equation is x√(1 + 2y) + y = x².
2. Apply the product rule to the term x√(1 + 2y). The product rule states that d(uv)/dx = u'v + uv', where u = x and v = √(1 + 2y).
3. Differentiate u = x to get u' = 1. For v = √(1 + 2y), use the chain rule: v' = (1/2)(1 + 2y)^(-1/2) * (2(dy/dx)).
4. Combine the derivatives using the product rule: d(x√(1 + 2y))/dx = 1 * √(1 + 2y) + x * (1/2)(1 + 2y)^(-1/2) * 2(dy/dx).
5. Differentiate the remaining terms: dy/dx for y and 2x for x². Set the derivatives equal: √(1 + 2y) + x(1 + 2y)^(-1/2)(dy/dx) + dy/dx = 2x. Solve for dy/dx and evaluate at P(1,0).

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 like x√(1 + 2y) + y = x², where y is not 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. In implicit differentiation, the chain rule is crucial when differentiating terms involving y, as y is considered a function of x.

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(1,0), into the derivative formula. This step is important for understanding the behavior of the function at specific locations on its graph.