Implicit+Differentiation

=I m p l i c i t D i f f e r e n <span style="color: #ffff00; font-family: 'Courier New',Courier,monospace; font-size: 190%;">t <span style="color: #00ff59; font-family: 'Courier New',Courier,monospace; font-size: 190%;">i <span style="color: #0101f9; font-family: 'Courier New',Courier,monospace; font-size: 190%;">a <span style="color: #6f549c; font-family: 'Courier New',Courier,monospace; font-size: 190%;">t <span style="color: #a423af; font-family: 'Courier New',Courier,monospace; font-size: 190%;">i <span style="color: #eb0000; font-family: 'Courier New',Courier,monospace; font-size: 190%;">o <span style="color: #ee9c6d; font-family: 'Courier New',Courier,monospace; font-size: 190%;">n = =media type="custom" key="5792925"= So. Implicit Differentiation is a funky, yet handy concept we use when we come across strange and outlandish equations like...

467 = 308x + 201xy + y 2 or... x 2 + 5x - 4y 2 + 3y + 21 = 0 == How the heck would you derive something like that? Can't do it, you say? Don't worry, if George can do it, you can too!! Quite simply, you derive implicitly. This means you derive both sides of the equation. When deriving a y-term, you use the chain rule because it is //implied// that y is a function of x. Let's do a relatively simple problem to start, shall we?

y 8 = x 5

8y 7 y' = 5x 4 ........ ...................Derive both sides of the equation remembering the chain rule.

y' = 5x 4 / 8y 7 .......................Solve for y'.

<span style="color: #ff00ff; font-family: 'Arial Black',Gadget,sans-serif; font-size: 160%;">__ANOTHER EXAMPLE!__ x 3 - 7x 2 y 3 + 4y 2 = -16

3x 2 - 14xy 3 + 7x 2 3y 2 y' + 8yy' = 0..........................derive implicitly, remembering chain rule and product rule

y'(7x 2 3y 2 + 8y) = -3x 2 + 14xy 3 .............................factor out a y' and move other variables to other side of equations

y' = (-3x 2 + 14xy 3 ) / (7x 2 3y 2 + 8y)........................solve for y'

A good application of this method would be in situations like this: Say you have the equation of a circle, for example //x// 2 + //y// 2 = 25, and you need to find the slope of the line tangent to the graph at the point (3, -4).

You could solve for y and derive explicitly, then use the negative square root (the point being on the bottom half of the circle). You would get a slope of 3/4. Or, you could derive implicitly. //D// ( //x// 2 ) + //D// ( //y// 2 ) = //D// ( 25) 2//x// + 2 //y y//' = 0 2 //y y//' = - 2 This method can be extremely useful considering you won't be able to derive explicitly in all situations. ** (Crazy, right??) ** It's ok though because you can do problems like this now:

Find the slope and concavity of the graph of //x//2//y// + //y//4 = 4 + 2//x// at the point (-1, 1).

D ( x 2 y + y 4 ) = D ( 4 + 2x) D ( x 2 y ) + D ( y 4 ) = D ( 4 ) + D ( 2x ) x 2 y' + 2xy + 4y 3 y' = 0 + 2 x 2 y' + 4y 3 y' = 2 - 2xy y'( x 2 + 4y 3 ) = 2 - 2xy

So at the point (1,-1): Then to finish the problem you can find the second derivative to determine its concavity. Since y''<0 at (-1,1), the graph is concave down.

Here is another handy application for implicit differentiation:



This is an example with some nice background music and a guy that draws weird x's:

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Thanks to.... http://archives.math.utk.edu/visual.calculus/3/implicit.7/4.html** http://www.math.ucdavis.ed u/~kouba/CalcOneDIRECTORY/ implicitdiffdirectory/Impl icitDiff.html
 * DONE! Now you can derive implicitly like Curious George! We will end with another cute fluffy animal.

Topic: Implicit Differentiation Group Members: Katie and Jen


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I love the intro and use of images and video to engage the audience. Nice accurate examples with graphs and applications to bolster the topics. Notation is good and steps are well explained. Could have used some more difficult examples that employ the product rule. Nice work!

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