This document summarizes a calculus lesson on implicit differentiation. The lesson began with examples of implied relationships in pictures and led into a discussion of implicit functions hidden within explicit functions. The class then learned how to take the derivative of an implicit function using the example of finding the derivative of any point on a circle. The process involves rewriting the equation in terms of y, applying the chain rule, and solving for the derivative of y. The lesson concluded by emphasizing that the derivative is zero when the numerator is zero, indicating a horizontal tangent line, and when the denominator is zero, indicating a vertical tangent line.

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Return of the Jedi
Yoda

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So, Some people (those who are not in our class, or
weren't there yesterday) reading this Scribe post may be a
little confused by the title.
The title is as such because Mr. K. actually started off the
class with a couple quot;Jedi mind tricksquot;!!! Isn't that
awesome?!?!
I won't go into detail, but let's just say:
THERE ARE NO GREY ELEPHANTS IN
DENMARK!!!
And so I begin the Calculus Related part of this Scribe post...

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Well, Mr. K. then shoed us a couple pictures (man I wish the
slides were up) with questions pertaining the existence of
somethings that were not shown in the pictures, but were
eluded to by certain hints in these pictures.
i.e. You see a shadow of person on the ground, but have not turned around
to see this person yet. Is there someone behind you???
When we gave an answer he asked us why? We said because it
is implied. There aren't any visuals of the objects in question,
but certain evidence allows us to imply that they exist.
This leads to the topic of the day:
IMPLICIT FUNCTIONS!!!

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So we moved on to a quick review of composite functions
and the chain rule, because we would have to use it later
on in the lesson.
In one example the answer was:

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Then we were asked to identify THIS function:

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It is, of course, a SEMI CIRCLE

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Now what's the equation of the full circle???

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And that can be drawn as...

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2
2
-2
-2

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However, a circle is not a function. But, is there a
function (besides √4-x2 and its opposite: -√4-x2) that is a
part of the circle???

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Sure there is: There's this one...
2
2
-2
-2

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And this one...
2
2
-2
-2

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And this one...
2
2
-2
-2

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Do you understand that there are an
infinite amount of functions with x as the
variable?
All of these functions hidden in the main
graph of a circle, are
IMPLICIT FUNCTIONS!!!

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Now think of a relation such as the circle given
previously, and think how you would find the derivative
of any given point in the domain that relation?
There are infinite implicit functions, so who knows which
one you should find the derivative of because there are
many of these functions that go through the same points!

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The only way to find the derivative of a
circle at any given point is by finding the
derivative of the relation using the
following derivation:

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First of all, let's use the following
equation of a circle:

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Now let's find the derivative:

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First rewrite it as such:

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Now we can recognize that the derivative
of 25 is always 0.

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Next, use the chain rule on the left side of
the equation...
remember it is:

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Identify ƒ(x), g(x), and their derivatives:
The reason why the
inner function quot;g(x)quot;
is simply quot;yquot; is
because the inner
function of y^2 is an
implicit function. We
do not know exactly
what it is, but we
know it is there.

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Did everyone see the red text on the
previous slide? It is very important!
The reason why the inner
function quot;g(x)quot; is simply
quot;yquot; is because the inner
function of y^2 is an
implicit function. We do
not know exactly what it
is, but we know it is there.

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Anyway, now using the outer and inner
functions, complete the chain rule:

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Now combine it with the derivative of x2
(2x) and you have the derivative of the left
side of the equation...

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therefore...

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Now solve for quot; y' quot;...

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So now we know that the derivative (slope
of the tangent line) at any given point on
the circle x2 + y2 = 25 is given by the
following formula

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So let's test it out.
Let's say we need to find the derivative of
the previous function at x=3.

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First, find the output of when x=3:

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The next step is simple. To find the derivative
at that point, plug the coordinates into the
equation of the derivative of the circle:

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The derivative (slope of the tangent) at x = 3
is (-3)/4... Let's see if that is true.

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So yes, the derivative (slope of the tangent) at
any point of any circle with its center at the
origin (because the derivative of the constant
will always be zero) is given by:

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From there, Mr. K. gave us some
examples to try and find the
derivative functions of relations
(which could have been seen on
the slides if they were up).

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Then one final point was made:
When the numerator of the
derivative function of the relation
is 0, the tangent is a horizontal
line.
When the denominator is 0, then
the tangent is a vertical line.
(as seen on the next slide)

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About it, that is...
For reading, thank you...
Grey-M, the next
scribe is...
Yoda