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Understanding Isolating Variables and Implicit Differentiation in Functions

This lesson explores the challenges of isolating variables in mathematical functions, particularly in the context of the relation y = x³ + x. We will discuss whether this relation is a function and whether its inverse qualifies as a function, supported by justification. Additionally, the tutorial addresses the difficulties of isolating the dependent variable but emphasizes the importance of finding the slope using the derivative. Examples will be provided, including implicit differentiation techniques for circles and comparative advantages of different methods.

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Understanding Isolating Variables and Implicit Differentiation in Functions

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  1. The trouble with isolating y... Is the relation y = x3 + x a function? Justify your answer. Is the inverse of the above relation a function? Again, justify! Graph: Equation: (Erase) Is y a function of x? Can we isolate the dependent variable?

  2. As you have seen on the previous slide, sometimes isolating the dependent variable of a function is difficult or even impossible. However, this should not prevent us from finding the slope of the function! (i.e. determining the derivative) Original equation Take the derivative of both sides Take the derivative of each term separately Use chain rule to take the derivative of y2 Common factoring Isolating the derivative

  3. Practice: Determine the slope of the function 2xy - y3 = 4 at the point (3, 2).

  4. Note: Implicit differentiation may be useful even in cases where it is possible to isolate the dependent variable For example, calculate the slope of the circle x2 + y2 = 25 at (3, -4): Explicit differentiation: Implicit differentiation: Is there an advantage to using the second method to solve this problem?

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