Name: Worked example: Derivative of ln(ÃÂx) using the chain rule | AP Calculus AB | Khan Academy
Uploaded: 2022-09-09T14:55:45.000Z
Duration: 6 min 45 s
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being equal to the natural log of the square root of x.

Past simple

And the key here is to recognize that f can actually

be viewed as a composition of two functions.

And we can diagram that out, what's going on here?

Well if you input an x into our function f,

So if we start off with some x, you input it,

the first thing that you do, you take the square root of it.

You are going to take the square root of the input

So then you take the natural log of that,

into another function that takes the natural log

Well you produce the natural log of the square root of x.

So you could view f of x as this entire set,

this combination of functions right over there.

then taking that output and inputting it into another.

which takes the square root of whatever its input is,

so u of x is equal to the square root of x.

and input it into another function that we could call v,

Well it take the natural log of whatever the input is.

In this case, in the case of f, or in the case of how

I just diagrammed it, v is taking the natural log,

the input happens to be square root of x,

so it outputs the natural log of the square root of x.

If we wanted to write v with x as an input,

we would just say well that's the natural log,

So that is v of the square root of x, or v of u of x.

So it is a composition which tells you that,

okay, if I'm trying to find the derivative here,

the chain rule is going to be very, very, very, very useful.

And the chain rule tells us that f prime of x

is going to be equal to the derivative of,

times the derivative of this inside function

Well, we know how to take the derivative of u of x

and v of x, u prime of x here, is going to be equal to,

well remember, square root of x is just the same thing

as x to 1/2 power, so we can use the power rule,

bring the 1/2 out from so it becomes 1/2 x to the,

and then take off one out of that exponent,

so that's 1/2 minus one is negative 1/2 power.

And what is v of x, sorry, what is v prime of x?

Well the derivative of the natural log of x

is one over x, we show that in other videos.

we know what v prime of x is, but what is v prime of u of x?

Well v prime of u of x, wherever we see the x,

we replace it, let me write that a little bit neater,

we replace that with a u of x, so v prime of u of x

This thing right over here, we have figured out,

and this thing, u prime of x, we figured out,

and x to the negative 1/2, I could rewrite that as 1/2

times one over x to the 1/2, which is the same thing

as 1/2 times one over the square root of x,

or I could write that as one over 2 square roots of x.

Well this is going to be equal to, in green,

v prime of u of x is one over the square root of x,

times, times, u prime of x is one over two times

the square root of x, now what is this going to be equal to?

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Worked example: Derivative of ln(ÃÂx) using the chain rule | AP Calculus AB | Khan Academy

entire

essentially

recognize

figure

Worked example: Derivative of ln(ÃÂx) using the chain rule | AP Calculus AB | Khan Academy

entire

essentially

recognize

figure

Worked example: Derivative of ln(ÃÂx) using the chain rule | AP Calculus AB | Khan Academy