We
began class by taking our quiz on IW’s 6 & 7. Once that was done we
reviewed and praised Noah’s scribe post; “That’s what an A+ scribe
post looks like” said O'Brien with much zeal.
The main topic of today’s class was ...
Derivative of the Inverse Function
Since we already know how to find the derivative of practically every other type of function, why not inverse functions too? (I CAN THINK OF PLENTY OF REASONS TO NEVER LEARN THIS, but my opinions are not influential in the slightest)
However before we could delve into derivatives of inverse functions OB reminded us that he had posted the solutions/answers to IW#8 : https://dl.dropbox.com/u/3243156/CHRHS/apcalc/U2%20ap%20IW%20%238%20solutions.pdf
We then encountered a problem with the solution to “Negative Integers” on Question 2 of IW #8, OB said he would change it.
* It was also hinted that #3 on IW #8 could possibly be a finely crafted opportunity day question *
OKAY SO , The majestic Inverse Function
We already know that to find the inverse of a function we replace f(x) with y and solve for 
. Easy peasy.
. Easy peasy.
(If you want a more in-depth refresher of finding the Inverse of a function patrickJMT is your BFF)
We also know that if you graph a function and its inverse, they have y = x symmetry. These are created by interchanging the
and 
values. For
example : 
If you were to graph these functions it would look something like
with the diagonal y = x symmetry. At any point on the function its
inverse point on the inverse is its reflection across the y = x line.
Example : if you have a point (1, 2) inverse is (2,1).
Algebraically : If f and g are inverses then :
and if we take the derivative with respect to we get :
we get : 
which then can be rewritten as:
Next, we looked at an example. Find:
Finding the derivative of this inverse by using the Chain Rule:
where
1 is the derivative of (x -1) in the Chain Rule.
Then we plug in 2 for x, since that was the original goal (finding the derivative of the inverse function) at
x = 2 :
and if we then solve we get a lovely small answer of :
Now, let's compare this answer to the answer we would get if we used the reciprocal relationship we found above.
If (2, 1) is a point on the inverse of f , then (1, 2) is a point on f.
So,
And if we plug in 1 for 
,we get something like
,we get something like 
and if we go back to the our first answer of
we can see that it was the reciprocal.
The key to this approach was comparing points on f and points on f'.
The key to this approach was comparing points on f and points on f'.
Next class we will learn how to find the derivative of the inverse function of
at x = 3
at x = 3
How exciting.
And as always continue to post questions!
To do for next class : IW #9
Keep working on IW #8 and older IW
p. 169/1, 2, 4
p. 171/Exploration 1
p. 175/29b & c
#APCalculusproblems
#Whydowedothistoourselves
#O'Brienlikestomakeuscry
UPDATE:
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