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Implicit Differentiation

Implicit Differentiation. By Dr. Julia Arnold and Ms. Karen Overman using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text. Up to this point, all of our functions have been expressed as

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Implicit Differentiation

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  1. Implicit Differentiation By Dr. Julia Arnold and Ms. Karen Overman using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text

  2. Up to this point, all of our functions have been expressed as y = f(x) , that is y is expressed explicitly in terms of the independent variable x. However, not all functions are expressed in this form. Consider for example, the equation: Since this equation is not solved for y we call this an implicit equation and we understand that y is a function of x. This equation can be solved for y as follows:

  3. Though you have seen it is possible to solve this equation for y, let’s use this simple example to start looking at taking the derivative implicitly by starting with the equation in its implicit form. Think of y as always requiring the Chain Rule since y is a function of the variable x. Thus the derivative of y is 1y ’, the outside derivative 1 times the inside derivative y ’. Taking the derivative implicitly you will need to use the Product Rule.

  4. After taking the derivative implicitly, it is helpful to solve for y’ and then if possible get y’ in terms of x.

  5. Now that we have found the derivative implicitly and found the derivative to be . Let’s compare with the derivative of , the original equation solved explicitly for y. The two derivatives are the SAME!!

  6. Now, let’s look at a slightly more complicated equation in implicit form. This equation can also be solved for y.

  7. Let’s repeat the same comparison in this problem. We’ll find the derivative implicitly and explicitly and compare the two. Keep in mind when taking the derivative implicitly you should think of y as always requiring the Chain Rule. Now solve for y’. Notice that the derivative is in terms of x and y not just x.

  8. In this example we could solve for y, , so now we can substitute this value of y in our derivative and find the derivative in terms of only x.

  9. Just for the sake of comparison, let’s find the derivative of the explicit equation, . You’ll notice this is the SAME as the derivative we found implicitly.

  10. Consider the following example. We would be hard pressed to solve this equation for y in order to obtain the derivative. As long as it is okay to leave the answer in x and y we can find using implicit differentiation. Using the Chain Rule on y means every derivative taken involving y should have y ‘ multiplied by it. Let’s find y ‘ for this:

  11. We will leave the final answer in the two variables x and y.

  12. Problem: Consider the equation . Find by implicit differentiation. Find the slope of the tangent line to the graph of the function y = f(x) at the point and find an equation of the tangent line at . Solution: Remember that this equation defines a circle centered at the origin with a radius of 2. Find by implicit differentiation.

  13. Solution Continued: Since y’= -x/y, the slope at is An equation of the tangent line at .

  14. Here is a graph of the circle and the tangent line to the circle at the point .

  15. In summary, when you are performing implicit differentiation you must treat the derivative of y as a Chain Rule since y is a function of x and multiply by y’.

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