(a) find two explicit functions by solving the equation for...

(a) find two explicit functions by solving the equation for $y$ in terms of $x,($ b) sketch the graph of the equation and label the parts differentiate the explicit functions, and (d) find $d y / d x$ implicitly and show that the result is equivalent to that of part (c). $16 y^{2}-x^{2}=16$

Key Concepts

Implicit Differentiation

Implicit differentiation is used when y is defined implicitly by an equation, instead of being isolated on one side. It requires differentiating both sides of the equation with respect to x, treating y as a function of x, and applying the chain rule appropriately. This method allows us to find the derivative dy/dx without first solving for y explicitly, and it must yield the same result as explicit differentiation for consistency.

Explicit Differentiation

This concept entails applying conventional differentiation rules to functions that are expressed explicitly, such as those involving square roots. Key techniques include the chain rule, which is necessary when differentiating compositions of functions. Explicit differentiation provides the slope of the curve for each branch of the function.

Branches of a Function

When an equation is solved for a squared variable, the process of taking square roots produces two distinct functions or branches. Each branch represents one possible set of output values (one positive and one negative) and is essential to fully capture the overall graph of the equation.

Graphing Conic Sections – Hyperbola

The given equation represents a conic section, specifically a hyperbola. Recognizing a hyperbola involves rewriting the equation into its standard form and identifying its key features such as vertices, axes, and asymptotes. Graphing the hyperbola accurately demands an understanding of how these elements interact and define the shape and position of the curve.

Solving for an Explicit Function

This concept involves algebraically manipulating an equation to isolate the variable y in terms of x. In cases where the equation involves y², solving for y typically requires taking square roots, which yields two solutions corresponding to the positive and negative roots. This lays the groundwork for understanding the behavior of the function in explicit form.

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