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U2S4V09 Differentiating Another Complicated Function.txt
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U2S4V09 Differentiating Another Complicated Function.txt
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#
# File: content-mit-18-01-1x-captions/U2S4V09 Differentiating Another Complicated Function.txt
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# Captions for MITx 18.01.1x module [70PoOJZpVOk]
#
# This file has 63 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.
#
#----------------------------------------
In our previous video, we talked about this function f.
And then in the problems, we gave you this function g.
And they look very similar.
But the ways in which you differentiate them
are different.
So for f, if you remember, we made this flowchart,
and we determined that the last thing
you would do when calculating the function is multiply.
So we started differentiating f using the product rule.
Now we're not going to make a flowchart for g.
That's a little bit too much work.
And we don't normally do that.
But we can still think about what
it would take to calculate g.
So you could take the tangent of your x.
And then you'd multiply by x.
And then the last thing that you do
is apply the square root function
or compose with the square root function.
And so that tells us that we can write g of x as u to the 1/2,
where u is just what's inside the square root.
It's x tangent x.
And so we're going to be using the chain rule to start off.
So our g prime is going to be what we need to differentiate
this 1/2 power on the u.
So using the power rule that's just 1/2 u to the minus 1/2.
But then we also need the derivative of u tagging along.
That's our chain rule second factor.
So putting everything back in terms of x, this
is 1/2 times the quantity x tangent x to the minus 1/2.
And then we need the derivative of x tangent x.
And for that, well, that we need the product rule
to differentiate.
So we're going to get our 1/2 times
x tangent x to the minus 1/2 times--
and then for the derivative of this,
we need the first function that's
x-- times the derivative of the second function.
So the derivative of tangent is secant
squared x plus the second function times the derivative
of the first function.
The first function is x, so it's derivative is 1.
And there we go.
We're done.
That's our derivative of g.
Now down the road you might encounter
some functions that are even more complicated than these.
But the thing to realize is that there's always
going to be a last step in computing any function.
And so either that last step will be arithmetic,
like taking a product or a quotient-- in which case,
you have a rule to work with that--
or you'll just be composing with some basic function.
And in that case, you have the chain rule.
So no matter what, you can always figure out how to get
started.
It's sort of like peeling an onion.
You just go one layer at a time.
So there's never any reason to be intimidated by a function.
So just get out there and take some derivatives.