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Studying for a test? Prepare with these 13 lessons on Derivative rules.

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# Worked example: Differentiating √(3x²-x) using the chain rule

Video transcript

What I want to do in this video
is start with the abstract-- actually, let me call it
formula for the chain rule, and then learn to apply it
in the concrete setting. So let's start off
with some function, some expression that
could be expressed as the composition
of two functions. So it can be expressed
as f of g of x. So it's a function that can
be expressed as a composition or expression that
can be expressed as a composition
of two functions. Let me get that same color. I want the colors
to be accurate. And my goal is to take the
derivative of this business, the derivative
with respect to x. And what the chain
rule tells us is that this is going to be
equal to the derivative of the outer function with
respect to the inner function. And we can write that as f
prime of not x, but f prime of g of x, of the
inner function. f prime of g of x
times the derivative of the inner function
with respect to x. Now this might seem all
very abstract and math-y. How do you actually apply it? Well, let's try it
with a real example. Let's say we were trying
to take the derivative of the square root of
3x squared minus x. So how could we
define an f and a g so this really is
the composition of f of x and g of x? Well, we could define f of x. If we defined f of x as being
equal to the square root of x, and if we defined g of x as
being equal to 3x squared minus x, then what
is f of g of x? Well, f of g of x is
going to be equal to-- I'm going to try to keep
all the colors accurate, hopefully it'll help
for the understanding. f of g of x is equal to--
where everywhere you see the x, you replace with the g of x--
the principal root of g of x, which is equal to the
principal root of-- we defined g of x right over
here-- 3x squared minus x. So this thing right
over here is exactly f of g of x if we define
f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule. What is f prime of g of
x going to be equal to, the derivative of f
with respect to g? Well, what's f prime of x? f prime of x is equal to--
this is the same thing as x to the 1/2 power, so we
can just apply the power rule. So it's going to be
1/2 times x to the-- and then we just take 1 away
from the exponent, 1/2 minus 1 is negative 1/2. And so what is f
prime of g of x? Well, wherever in the
derivative we saw an x, we can replace it with a g of x. So it's going to be
1/2 times-- instead of an x to the negative 1/2, we
can write a g of x to the 1/2. And this is just going
to be equal to-- let me write it right over here. It's going to be
equal to 1/2 times all of this business to
the negative 1/2 power. So 3x squared minus x,
which is exactly what we need to solve right over
here. f prime of g of x is equal to this. So this part right
over here I will-- let me square it off in green. What we're trying to
solve right over here, f prime of g of x,
we've just figured out is exactly this thing
right over here. So the derivative of f of the
outer function with respect to the inner function. So let me write it. It is equal to 1/2 times g
of x to the negative 1/2, times 3x squared minus x. This is exactly this
based on how we've defined f of x and how we've
defined g of x. Conceptually, if
you're just looking at this, the derivative
of the outer thing, you're taking something
to the 1/2 power. So the derivative
of that whole thing with respect to your something
is going to be 1/2 times that something to the
negative 1/2 power. That's essentially
what we're saying. But now we have to take the
derivative of our something with respect to x. And that's more straightforward. g prime of x-- we just use
the power rule for each of these terms-- is
equal to 6x to the first, or just 6x minus 1. So this part right over here
is just going to be 6x minus 1. Just to be clear,
this right over here is this right over here
and we're multiplying. And we're done. We have just applied
the power rule. So just to review,
it's the derivative of the outer function
with respect to the inner. So instead of having
1/2x to the negative 1/2, it's 1/2 g of x to
the negative 1/2, times the derivative of the
inner function with respect to x, times the derivative
of g with respect to x, which is right over there.