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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 2

Lesson 25: Radical functions differentiation# Worked example: Derivative of ∜(x³+4x²+7) using the chain rule

AP.CALC:

FUN‑3 (EU)

, FUN‑3.C (LO)

, FUN‑3.C.1 (EK)

Sal differentiates ∜(x³+4x²+7) and evaluates the derivative at x=-3.

## Want to join the conversation?

- I think that Sal left out the final step when solving for f'(-3). He finishes with an answer of 3/32 but neglects to multiply this by 3 (from u'(x)). I believe that the correct answer for f'(-3) is 9/32. No? Hopefully this solution wasn't critical to the programming of a manned orbiter, or those astronauts might be in for a fun ride!(2 votes)
- Actually, he multiplied the 3 (from u'(x)) with the 1/4, and thus became 3/4.

So 3/32 is the correct answer(25 votes)

- Shouldn't you be able to distribute the radical and then use the power rule?

f(x) = x^3/4 + 4x^2/4 + 7^1/4

f'(x) = 3/4x^(-1/4) + 2x^(-1/2) + 0(1 vote)- Radicals can't be distributed.

For example: (x + y) ^ 2 = x^2 + 2xy + y^2

This can't be written as x^2 + y^2(18 votes)

- at0:13he says we are working with a composite function, how do we know if we are or not?(4 votes)
- Basically it means if you have a function inside of a function. In the title of the video you have the polynomial starting with x^3 inside of the fourth root. This can also take the form of something like sin(ln(x)) with ln being inside of sin or something like e^sqrt(x) where the square root function is "inside" of e^x. It's a little tricky to spot at first but you get the hang of it with practice.

It may help if you imagine that you are able to replace one part of the function with a g(x) where g(x) equals that function. Let me know if that didn't help.(3 votes)

- When Sal gets to the derived function (before evaluating for -3), can that be simplified any more? It seems too complex or messy.(2 votes)
- Not really. The part raised to the -3/4 power could be rewritten as 1/(the fourth root of some polynomial), but nothing would be able to cancel out.(2 votes)

- A little bit contradicts or I have misunderstood: We know that derivative of X^1/2 is (1/2)*X^(1/2-1). But now, according to this exercise this formula changes to the: [ (1/2)*X^(1/2-1) ]* Derivative of X.(2 votes)
- The problem does not want us to find the derivative of x^(1/4). It wants us to find (x^3+4x^2+7)^(1/4). This is a COMPOSITION of functions, so the Chain Rule is needed. Because of this, we need to multiply by the derivative of the inner function. See Khan Academy's videos on the Chain Rule for additional help.(2 votes)

## Video transcript

- [Voiceover] Let's see if
we can take the derivative with respect to x of the fourth root of x to the third power plus
four x squared plus seven. And at first you might say, "All right, "how do I take the
derivative of the fourth root "of something, it looks like
I have a composite function, "I'm taking the fourth root
of another expression?" And you'd be right. And if you're dealing
with a composite function, the chain rule should be front of mind. But first, let's just
make this fourth root a little bit more tractable for us. And just realize that this fourth root is nothing but a fractional exponent. So this is the same
thing as the derivative with respect to x of x to
the third plus four x squared plus seven to the 1/4 power, to the 1/4 power. Now, how do we take
the derivative of this? Well, we can view this, as
I said a few seconds ago, we can view this as a composite function. What do we do first with our x? Well, we do all of this business, and we can call this u of x. And then whatever we get for u of x, we raise that to the fourth power. So the way that we would
take the derivative, we would take the derivative of this, you could view it as the outer function with respect to u of x
and then multiply that times the derivative
of u with respect to x. So let's do that. So what this is going to be, this is going to be equal to, so we're gonna take our outside function, which I'm highlighting in green now, so, or I take something to the 1/4, I'm gonna take the derivative
of that with respect to the inside, with respect to u of x. Well, I'm just gonna
use the power rule here. I'm just gonna bring that 1/4 out front, so it's gonna be 1/4 times whatever I'm taking the
derivative with respect to, to the 1/4 minus one power. Look, all I did was use
the power rule here. I didn't have an x here. Now I'm taking the derivative
with respect to u of x, with respect to this
polynomial expression here. So I could just throw the
u of x in here if I like, actually let me just do that. So, this is going to be x to the third plus four x squared plus seven. And then I wanna multiply that, and this is the chain rule. I took the derivative of the outside with respect to the inside and
then I'm gonna multiply that times the derivative of the inside. So what's the derivative of u of x? U prime of x, let's see we just gonna use the power rule a bunch of times, it's gonna be three x squared plus two times four is eight x to the two minus one is
just one power, first power, so I can just write that as eight x, and then the derivative
with respect to x of seven, well, the derivative with
respect to x of a constant is just gonna be zero. So that's u prime of x. So then I'm just gonna
multiply by u prime of x which is three x squared plus, three x squared plus eight x. And so, well, I can clean
this up a little bit, so this would be equal to, this would be equal to. Actually let me just
rewrite that exponent there. So this 1/4 minus one, I can rewrite it, 1/4 minus one is negative 3/4, negative 3/4, negative 3/4 power. And you can manipulate this in
different ways, if you like, but the key is to just
recognize that this is an application of the chain rule. Derivative of the outside, well, actually, the first thing to
realize is the fourth root is the same thing as taking
something to the 1/4 power, basic exponent property, and then realize, okay, I have a composite function here. So I can take the
derivative of the outside with respect to the inside,
that's what we did here, times the derivative of the
inside with respect to x. And so if someone were to tell you, if someone were to say,
"All right, f of x, "f of x is equal to the fourth root of "x to the third plus four
x squared plus seven," and then they said,
"Well, what is f prime of "I don't know, negative three?" Well, you would evaluate
this at negative three. So let me just do that. So it's 1/4 times, see, you have negative 27, I hope this works out reasonably well, plus 36, plus 36, plus 7 to the negative 3/4, what does this result to? This is going to be equal to, this right over here is 16, right? Negative 27 plus seven is negative 20 plus 36, so this is 16. I think this is going to work out nicely. And then times three times negative, so three times nine, which is 27 minus 24. So this is going to be right over here, that is going to be three. Now, what is 16 to the negative 3/4? So this is 1/4 times,
so 16 times to the 1/4 is two and then you raise that to the, let me, actually I don't
want to skip steps here. At this point we're dealing with algebra, or maybe even pre-algebra. So this is going to be times, times, 16 to the 1/4, and then we're gonna raise
that to the negative three times that three out front. So we could put that three there. 16 to the 1/4 is two, two to the third is eight, so two to the negative third power is 1/8, so that is 1/8. So we have 3/4 times 1/8 which is equal to three over 32, 3/32. So that would be the
slope of the tangent line of the graph y is equal to f of x when x is equal to negative three.