Radical functions differentiation | Derivative rules | AP Calculus AB | Khan Academy

Let's see if we can take the derivative with respect to (x) of the fourth root of (x^3 + 4x^2 + 7). At first, you might say, "All right, how do I take the derivative of a fourth root of something?" It looks like I have a composite function; I'm taking the fourth root of another expression here. And you'd be right!

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 really nothing but a fractional exponent. So, this is the same thing as the derivative with respect to (x) of ( (x^3 + 4x^2 + 7)^{1/4} ).

Now, how do we take the derivative of this? Well, we can view this, as I said a few seconds ago, as a composite function. What do we do first with our (x)? Well, we do all of this business, and we could call this (U) (or (U(x))). Then, whatever we get for (U(x)), we raise that to the fourth power.

The way that we would take the derivative is we would take the derivative of this, which you could view as the outer function with respect to (U(x)), and then multiply that times the derivative of (U) with respect to (x). So, let's do that!

This is going to be equal to... So, we are going to take our outside function, which I'm highlighting in green now. So where I take something to the (1/4), I'm going to take the derivative of that with respect to the inside (with respect to (U(x))). Well, I'm just going to use the power rule here; I'm just going to bring that (1/4) out front. So it's going to be ( \frac{1}{4} ) times whatever I'm taking the derivative with respect to raised to the ( \frac{1}{4 - 1} ) power.

Look, all I did is use the power rule here. I didn't have an (x) here. Now I'm taking the derivative with respect to (U(x)), with respect to this polynomial expression here. So I could just throw the (U(x)) in here if I like. Actually, let me just do that. So this is going to be ( (x^3 + 4x^2 + 7)^{1/4} ) and then I want to multiply that. And this is the chain rule; I took the derivative of the outside with respect to the inside, and I'm going to multiply that times the derivative of the inside.

So what's the derivative of (U(x)) or (U')? Let's see, we’re just going to use the power rule a bunch of times. It's going to be (3x^2 + 2 \cdot 4x^{2 - 1}), which is just (8x). And then the derivative with respect to (x) of seven (well, the derivative with respect to (x) of a constant) is just going to be zero. So that's (U'(x)).

So then I'm just going to multiply by (U'(x)), which is (3x^2 + 8x).

I can clean this up a little bit. So this would be equal to, actually, let me just rewrite that exponent there. So (1/4 - 1) I can rewrite as (-\frac{3}{4}) power. And you could manipulate this in different ways if you like, but the key is to just recognize that this is an application of the chain rule: 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). So if someone were to tell you, "All right, (f(x) = \frac{1}{4}) root of (x^3 + 4x^2 + 7)," and then they said, "Well, what is (f’(-3))?" Well, you would evaluate this at (-3).

So let me just do that. So it's (\frac{1}{4} \times (-27 + 36 + 7)^{-\frac{3}{4}}). What does this result to? This right over here is (16). Right? So (-27 + 7) is (-20) plus (36), so this is (16).

I think this is going to work out nicely. Then times (3 \times (-3)^2), which is (3 \cdot 9) which is (27), minus (24). So this is going to be... right over here... that is going to be (3).

Now, what is (16^{-\frac{3}{4}})? So let me... (= \frac{1}{4}). So (16^{\frac{1}{4}} = 2), and then you raise that to... let me, actually, I don't want to skip steps here, but at this point we are dealing with algebra or maybe even pre-algebra.

So this is going to be times (16^{\frac{1}{4}}), and then we’re going to raise that to the (-3) times that three out front. So we could put that three there. (16^{\frac{1}{4}}) is (2). (2^3) is (8). So (2^{-\frac{3}{4}} = \frac{1}{8}).

So we have (3/4) times ( \frac{1}{8}), which is equal to ( \frac{3}{32} ).

So that would be the slope of the tangent line of the graph (y = f(x)) when (x = -3).