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# Pochodna złożenia trzech funkcji

Transkrypcja filmu video (w języku angielskim)

So now we're going
to attempt to take the derivative of the sine of
the natural log of x squared. So now we have a function that's
the composite of a function, that's a composite
of another function. So one way you
could think of it, if you set f of x as being equal
to sine of x, and g of x being the natural log of x, and
h of x equaling x squared. Then this thing right over
here is the exact same thing as trying to take the
derivative with respect to x of f of g of h of x. And what I want to
do is kind of think about how I would
do it in my head, without having to write all
the chain rule notation. So the way I would
think about this, if I were doing this in
my head, is the derivative of this outer function
of f, with respect to the level of composition
directly below it. So the derivative of
sine of x is cosine of x. But instead of it
being a cosine of x, it's going to be cosine of
whatever was inside of here. So it's going to be
cosine of natural log-- let me write that
in that same color-- cosine of natural
log of x squared. I'm going to do x that
same yellow color. And so you could really view
this, this part what I just read over here, as f prime,
this is f prime of g of h of x. This is f prime of g of h of x. If you want to keep
track of things. So I just took the derivative
of the outer with respect to whatever was inside of it. And now I have to
take the derivative of the inside with respect to x. But now we have another
composite function. So we're going to
multiply this times, we're going to do the
chain rule again. We're going to take the
derivative of ln with respect to x squared. So the derivative
of ln of x is 1/x. But now we're going
to have 1 over not x, but 1 over x squared. So to be clear, this part right
over here is g prime of not x. If it was g prime of x,
this would be 1 over x. But instead of an x, we
have our h of x there. We have our x squared. So it's g prime of x squared. And then finally, we
can take the derivative of our inner function. Let me write it. So we could write this
is g prime of h of x. And finally, we
just have to take the derivative of our innermost
function with respect to x. So the derivative of
2x with respect to-- or the derivative of x squared
with respect to x is 2x. So times h prime of x. Let me make everything clear. So what we have right
over here in purple, this, this, and this
are the same things. One expressed concretely,
one expressed abstractly. This, this, and this
are the same thing, expressed concretely
and abstractly. And then finally, this and
this are the same thing, expressed concretely
and abstractly. But then we're done. All we have to do to be done
is to just simplify this. So if we just change the order
in which we're multiplying, we have 2x over x squared. So I can cancel some out. So this x over x,
2x over x squared is the same thing as 2 over x. And we're multiplying it
times all of this business. So we're left with 2 over x. And this goes away. 2 over x times the cosine of
the natural log of x squared. So it seemed like a very
daunting derivative. But we just say, OK, what's the
derivative of sine of something with respect to that something? Well, that's cosine
of that something. And then we go in
one layer, what's the derivative of
that something? Well, in that something we
have another composition. So the derivative of ln of x,
or ln of something with respect to another something,
well that's going to be 1 over
the something. So we had gotten a 1
over x squared here, that squared got canceled out. And then finally, the derivative
of this innermost function, it's kind of like
peeling an onion. The derivative of
this inner function with respect to x,
which was just 2x. Which we got right over here. This was 1 over x squared. This was 2x before we
did any canceling out. So hopefully that helps
clear things up a little bit.