Implicit differentiation
0 likes1,403 views
To summarize: 1) To find the derivative of an implicit function y=y(x) defined by an equation F(x,y)=0, take the derivative of both sides with respect to x. 2) This will give a new equation involving x, y, and dy/dx that can be solved for dy/dx. 3) The examples show applying this process to find derivatives and tangent lines for various implicit equations.
More Related Content
![Week 3 [compatibility mode]](https://cdn.slidesharecdn.com/ss_thumbnails/week3compatibilitymode-130213164415-phpapp02-thumbnail.jpg?width=560&fit=bounds)
Similar to Implicit differentiation (20)
![Lecture 1a [compatibility mode]](https://cdn.slidesharecdn.com/ss_thumbnails/lecture1acompatibilitymode-130523045820-phpapp02-thumbnail.jpg?width=560&fit=bounds)
Implicit differentiation
- 1. Compute derivatives of implicit functions Facts: An equation F (x, y) = 0 involving variables x and y ( may define y as a func- dy tion y = y(x). To compute y = dx , one can apply the following procedure. (Step 1) View y = y(x) and differentiate both sides of the equation F (x, y) = 0 with respect to x. This will yield a new equation involving x, y and y . (Step 2) Solve the resulting equation from (Step 1) for y . dy Example 1 Given x4 + x2 y 2 + y 4 = 48, find dx . Solution: View y = y(x) and differentiate both sides of the equation x4 + x2 y 2 + y 4 = 48 to get 4x3 + 2xy 2 + 2x2 yy + 4y 3 y = 0. To solve this new equation for y , we first combine those terms involving y , (2x2 y + 4y 3 )y = −4x3 − 2xy 2 , and then solve for y : −4x3 − 2xy 2 y = . 2x2 y + 4y 3 Example 2 Find an equation of line tangent to the curve xy 2 + x2 y = 2 at the point (1, −2). dy Solution: The slope m of this line, is dx at (1, −2), and so we need to find y first. Apply implicit differentiation. We differentiate both sides of the equation xy 2 + x2 y = 2 with respect to x (view y = y(x) in the process) to get y 2 + 2xyy + 2xy + x2 y = 0. Then we solve for y . First we have (2xy + x2 )y = −y 2 − 2xy, and then −y 2 − 2xy y = . 2xy + x2 −(−2)2 −2(1)(−2) At (1, −2), we substitute x = 1 and y = −2 in y to get the slope m = 2(1)(−2)+12 = 0, and so the tangent line is y = −2. 1
- 2. Example 3 Find all the points on the graph of x2 + y 2 = 4x + 4y at which the tangent line is horizontal. Solution: First find y . We differentiate both sides of the equation x2 + y 2 = 4x + 4y with respect to x (view y = y(x) in the process) to get 2x + 2yy = 4 + 4y . Then we solve for y . First we have (2y − 4)y = 4 − 2x, and then 2−x y = . y−2 Note that when x = 2, the equation x2 + y 2 = 4x + 4y becomes 4 + y 2 = 8 + 4y, or √ √ y 2 − 4y = 4. Solve this equation we get y = 2 + 8 and y = 2 − 8. Therefore, at √ √ (2, 2 − 8) and (2, 2 + 8), the curve has horizontal tangent lines. 2