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U1S2V15 The existence of tangent lines and derivatives.txt
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U1S2V15 The existence of tangent lines and derivatives.txt
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# File: content-mit-18-01-1x-captions/U1S2V15 The existence of tangent lines and derivatives.txt
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# Captions for MITx 18.01.1x module [2Ms02JiZ6vQ]
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# This file has 72 caption lines.
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# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
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#----------------------------------------
You just did some problems where there
was no tangent line at a particular point, which
meant that there was no derivative at that point.
The two examples that we looked at
were the absolute value function,
which has a corner at the point x equals 0,
and the Heaviside function, which
has a jump discontinuity at the point x equals 0.
Let's look at these functions in a little bit more detail,
starting with the absolute value function.
Notice that there is no tangent line at x equals 0.
However, if we cover up the left-hand side
of this function, then at x equals 0
on this part of the function, there
appears to be a limit of secant lines approaching
from the right that is equal to this line itself.
So what does that mean, in terms of the derivative?
In terms of the derivative, this means
that this limit f of x minus f of 0
over x minus 0-- the definition of the derivative-- this limit
exists from the right.
As it's a right-sided limit, we call
this a right-sided derivative at 0,
and denote it as f prime of 0 plus,
to show that it exists for points to the right of 0.
What is this value?
In this case, it's the slope of this line, which is 1.
Now, if instead I cover the right-hand side of this graph,
I see that the graph has a limit of secant lines approaching
from the left.
That is, this left-sided limit gives us
a left-sided derivative, which is equal to negative 1.
Now, both the right and the left-sided derivatives exist.
They're not equal, so the derivative
does not exist at x equals 0.
Now, let's look at the heavy side
function, which has a jump discontinuity at x equals 0.
If I cover the left-hand side of this function,
because the function is equal to 1 at x equals 0,
this function does have a right-sided derivative,
which is equal to 0.
However, when we cover the right-hand side of this graph,
what we see is that because the function is equal to 0 almost
everywhere except at x equals 0, where the function is
equal to 1, the limit of secant lines from the left
approaches a vertical line.
So the left-sided derivative is infinite.
So once again, the right and left limits do not agree,
and so there is no derivative at x equals 0.
So these were two cases where the tangent line did not
exist at a point, and so the derivative did not
exist at that point.
Now my question is, if the tangent line exists at a point,
does the derivative exist there?
The answer is almost always yes, except for the case
of vertical tangent lines.
An example of this is the function, the cube root of x.
The graph of this function looks like so,
and this function has a well-defined tangent
line at x equals 0.
However, it is a vertical line.
The slope of a vertical line is infinite,
so there is no derivative at x equals 0.
Now that we've fully explored the relationship
between the existence of tangent lines
and the existence of derivatives,
I want you to do some practice problems to get more practice
estimating derivatives and determining
where derivatives do not exist.