1. 15. Implicit Differentiation Objectives: Refs: B&Z 10.4. 1.Learn some new techniques 2.Examples 3.An application

2. The equation is written in such a way, that given any value of x, we can directly calculate the corresponding y value. Consider the equation The equation clearly identifies y as a function of x; f(x). we have an explicit rule for f(x). In such an example, we can directly calculate the derivative, f ’ (x) of f with respect to x.

3. Here the rule for calculating y is not explicitly stated. However given a value for x we can still determine the corresponding y value. Now consider the equation For example, when x=2, 2y=2  y=1.

4. In fact both of these equations contain the same information. The difference is that in the second version, y is given implicitly as a function of x. One solution would be to rewrite the equation in terms of y, and then differentiate. We did this already! So if we wanted to calculatedy dx given xy=2 how would we do it?

5. How can we rewrite it so that y= f(x)? The answer is we can’t! The important thing to remember at all times is that y is still a function of x - it is not a constant. What about this equation: dy dx So if we are asked to find in such a case, we need to learn a technique to deal with it - implicit differentiation.

6. Example 1: dx dy Let xy=2. Calculate without rearranging the equation. First differentiate both sides of the equation with respect to x. d dx (xy)= d dx (2) this is a product this is a constant Now rearrange the equation

7. Note that y=2/x so that as before.

8. Example 2: d dx (x 3 +x+y 5 -2y)= d (3) dx dy Findwhenx 3 +x+y 5 -2y=3. Here we are differentiating y 5 with respect to x.

9. Remember that y is some function of x : f(x)=y So when we differentiate we have to use the chain rule. or

10. Returning to our problem: Now we simply rearrange the equation: We have

11. Example 3: (log e y+x 3 +y)= d (3) dx d dx dy Find when log e (y)+x 3 +y=3. Solution:

12. Application example. Find the slope of the curve x 2 + y 2 = 25 at the point (3,4). Solution (3,4) This is a circle of radius 5 We want the slope of the tangent at the point (3,4)

13. We need to rewrite the equation in terms of y. Direct method. (3,4) We want the slope at (3,4) - here y > 0 so we choose the positive square root. Let u(x)=25-x 2, so y=  u,

14. At the point x=3, Direct method (continued). (3,4) So

15. Implicit differentiation. (3,4) At the point (3,4)