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U2S4V09 Differentiating a Complicated Function.txt
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U2S4V09 Differentiating a Complicated Function.txt
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#
# File: content-mit-18-01-1x-captions/U2S4V09 Differentiating a Complicated Function.txt
#
# Captions for MITx 18.01.1x module [9karbu3LFEg]
#
# This file has 100 caption lines.
#
# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
#
#----------------------------------------
Let's say we have a function like this.
If we want to do stuff with this function,
like take its derivative, where would we even begin?
There's a lot of ingredients to this function.
So there's a square root, there's a trig function.
You're multiplying them.
There's an x squared in there somewhere.
It's pretty complicated.
How should we organize our thoughts about this function?
Something that I find helpful is to just think
about what you would do in order to compute f.
And so notice, I'm not talking about computing f prime right
now.
I'm just talking about computing the function itself.
So if you have a specific value for x
and you want to calculate f of x, what steps would you take?
So you might start by calculating the square root.
And I'll write that as x to the 1/2
since we're going to use the power rule on it down the road.
What else would you do?
You might calculate x squared.
That's something that we're going to need.
That's another power.
And then you're going to have to take
the tangent of the x squared.
So that's composing with the tangent function.
We're taking tangent of the result of x squared.
And so once we have these two parts,
we can just put them together, and we'll
get x to the 1/2 times tangent of x squared.
And that's our final function f of x.
So this last step, we're just multiplying these two things
together.
So that's our flow chart for how we would calculate f of x.
How does that help us?
Well, notice the last thing that we ever
do in computing f is we multiply these two parts.
So that's going to tell was that we
can write f of x as a product g of x times h of x, where g of x
is this x to the 1/2, and h of x is this tangent of x squared
over here.
So now if we want to differentiate f,
we know how to start because we have
f written as the product of two simpler functions.
So we can use the product rule.
f prime is just going to be the first function
times the derivative of the second
plus the second function times the derivative of the first.
Now, we're still going to have some work
to do to calculate h prime and g prime.
And it might not be straightforward to do that.
But the point is that h and g are simpler functions than f.
And so by analyzing f and realizing that, hey,
it takes this product form, that allows
us to reduce the problem to the simpler functions
h and g that come together in this way using the product
rule.
So let's keep going.
We have g of x being x to the 1/2.
And then h prime, that means we have
to take the derivative of tangent of x squared.
Next up is h of x, and that's tangent of x squared, times
the derivative of g.
So that's x to the 1/2 prime.
So here we've got x to the 1/2 times--
and then we need the derivative of this tangent of x squared.
Now if it was tangent of x, we'd know
exactly what to do because that's
one of our basic functions.
But instead, it's tangent of x squared.
So we can substitute a u for the x squared.
And we're using the chain rule here.
So we need the derivative of the tangent function,
and that's secant squared, applied to our u.
And then we need the derivative of u tagging along.
So that's 2x.
And then for our next term we have the tangent of x squared.
And then for the derivative of x to the 1/2,
that is just the power rule.
So we're going to get 1/2 times x to the 1/2 minus 1.
And we're done with all of our derivatives.
The rest is just algebra.
So we're going to get 2-- and then putting
all of the powers of x together, x to the 3/2 times
the secant squared of x squared.
And then for the next term, I'm going to put everything
into a fraction.
So we have tangent of x squared over 2 times x to the 1/2.
And we're done, that as our derivative.
So next up, we're going to differentiate a function that
looks very similar to this one.
But first we want you to think about how
we ought to start it in basically the same way
that we thought about this function.
So take a moment and look at that
and we'll come back very soon.