Lecture 11 implicit differentiation - section 3.5
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This document discusses implicit differentiation and finding the tangent line to a circle implicitly defined by an equation. It provides an example of using implicit differentiation to find the slope of the tangent line to the circle x^2 + y^2 = 25 at the point (3,4). The steps are to differentiate both sides of the equation with respect to x, solve for dy/dx, and plug in the x and y values of the point to find the slope of the tangent line.
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Lecture 11 implicit differentiation - section 3.5
- 2. 2 Circles Figure 1
- 3. Example 1 – Finding a Tangent Line Implicitly Implicit Solution: (a) Differentiate both sides of the equation x2 + y2 = 25: 3
- 4. 4 Example 1 – Solution Remembering that y is a function of x and using the Chain Rule, we have Thus Now we solve this equation for dy/dx: cont’d
- 5. 5 Example 1 – Solution At the point (3, 4) we have x = 3 and y = 4, so cont’d
- 6. 6 Implicit Differentiation Differentiate term-by-term Terms with a “y” (dependent variable) get a y’ Collect y’ terms on one side Factor out y’ Solve for y’ by dividing
- 7. 7 Implicit Differentiation - Example