Yes, this is homework. I’m just trying to figure it out, but if you have a prolem helping me, fine. It’s college work and there are no rules against getting help.
Basically, we’re working with derivatives in calc.
I need to find f’ (f-prime) of [f(x)]^3.
I did previously find f’ (f-prime) of [f(x)]^2: it’s 2 x f’(x) x f(x).
And [f(x)]^3 = [f(x)]^2 x [f(x)]
Is the answer 3 x f"(x) x f’(x) x f(x) ?
No, it isn’t. As long as you know the chain rule, you shouldn’t need to know the derivative of [f(x)]^2 to find the derivative of [f(x)]^3.
The chain rule says that:
if f(x) = g(h(x))
then f’(x) = h’(x) * g’(h(x))
(by the way, you probably shouldn’t use x as both your variable and as a multiplication sign – it’s confusing. The * is often used to indicate multiplication on computers).
We haven’t gotten to chain rule yet. That’s next.
Oh, ok. Then you can use the product rule:
if f(x) = g(x) * h(x)
f’(x) = g’(x) * h(x) + g(x) * h’(x)
Now you do need to know the derivative of [f(x)]^2, because as you say, [f(x)]^3 = f(x) * [f(x)]^2
What did you do to find the derivative of [f(x)]^2?
Presumably, whatever it was, you can do the same thing here.
Incidentally, referring to things as “f’ of [f(x)]^2” is very odd. In all the uses I’ve ever seen, f’ typically always means “The derivative of f”, not “The derivative of some arbitrary function built up from f”
Yep, if you haven’t quite gotten to the Chain Rule yet, I’m pretty sure Rysto’s (second) post describes the approach you need to take.
And I agree with Indistinguishable’s objection to your terminology. I’m pretty sure what you mean to say is “I need to find the derivative of…” not “I need to find f-prime of…” f’ (f-prime) means the derivative of f specifically. (Then f-prime of something would mean that you take the derivative of f and “plug in” the something in place of x.) Kind of a nitpick, but you’re less likely to confuse people (including yourself) if you use the proper terminology.
[QUOTE=Rysto]
Oh, ok. Then you can use the product rule:
if f(x) = g(x) * h(x)
f’(x) = g’(x) * h(x) + g(x) * h’(x)
Now you do need to know the derivative of [f(x)]^2, because as you say, [f(x)]^3 = f(x) * [f(x)]^2
[/QUOTE]
Sigh
Here’s how the problem is actually phrased.
Use your answer from blah blah blah* to help you find d/dx [f(x)]^3. Use the fact that [f(x)^2 = {f(x)]^2} * [f(x)]
*The answer is f’(x) [f(x)]^3 = 2 * [f’(x)] * [f(x)], and yes, that’s the format they want.
The product rule would be to multiply f’ by 2 * [f’(x)] * [f(x)], and then add the {derivative of 2 * [f’(x)] * [f(x)]} multipled by f(x) It’s that last bit which get’s me. I can use the shortcut rules or the longcut rules for less complex terms, but I can’t figure out how to get the short derivative of 2 * [f’(x)] * [f(x)] that I can use to multiply.
Sorry everyone. Maybe I’m just stupid, but the problem and the way they want me to doscover a “shortcut” doesn’t make any sense to me. Obviously, this is anticipation of teaching the chainrule.
I’m really not sure what they want from you here. The obvious approach using the product rule would be to say that:
a(x) = g(x) * h(x) where g(x) = f(x) and h(x) = [f(x)]^2
so a’(x) = g’(x) * h(x) + g(x) * h’(x)
I don’t see how that other fact mentioned in the problem is useful here.
[QUOTE=smiling bandit]
The product rule would be to multiply f’ by 2 * [f’(x)] * [f(x)], and then add the {derivative of 2 * [f’(x)] * [f(x)]} multipled by f(x) It’s that last bit which get’s me.
[/QUOTE]
[f(x)]^3 = [f(x)]^2 * [f(x)].
To use the Product Rule on that, each of those two things (the [f(x)]^2 and the [f(x)]) gets multiplied by the derivative of the other. So you just need to know the derivative of [f(x)]^2 (which you gave in your OP) and the derivative of f(x) (which is f’(x)).
[QUOTE=smiling bandit]
Sigh
Here’s how the problem is actually phrased.
Use your answer from blah blah blah* to help you find d/dx [f(x)]^3. Use the fact that [f(x)^2 = {f(x)]^2} * [f(x)]
*The answer is f’(x) [f(x)]^3 = 2 * [f’(x)] * [f(x)], and yes, that’s the format they want.
The product rule would be to multiply f’ by 2 * [f’(x)] * [f(x)], and then add the {derivative of 2 * [f’(x)] * [f(x)]} multipled by f(x) It’s that last bit which get’s me. I can use the shortcut rules or the longcut rules for less complex terms, but I can’t figure out how to get the short derivative of 2 * [f’(x)] * [f(x)] that I can use to multiply.
Sorry everyone. Maybe I’m just stupid, but the problem and the way they want me to doscover a “shortcut” doesn’t make any sense to me. Obviously, this is anticipation of teaching the chainrule.
[/QUOTE]
This isn’t anticipation of the chain rule. This is elaboration on the usefulness of the product rule.
As other have said, you’re beating the terminology to within an inch of its life here. You’re trying to find the derivative of f(x)[sup]3[/sup]. That is, [sup]d[/sup]/[sub]dx[/sub] f(x)[sup]3[/sup]. The phrase “f-prime of f cubed” doesn’t really mean anything, and it certainly doesn’t mean what you want it to mean.
This question can be answered using the chain rule, but it is in no way necessary. What you know is that f(x)[sup]3[/sup] = f(x)*f(x)[sup]2[/sup]. That is, f(x)[sup]3[/sup] is a product of two functions. You know how to take the derivative of a product of two functions, namely:
[sup]d[/sup]/[sub]dx[/sub] [f(x)f(x)[sup]2[/sup]] = f(x) [sup]d[/sup]/[sub]dx[/sub]f(x)[sup]2[/sup] + f(x)[sup]2[/sup] [sup]d[/sup]/[sub]dx[/sub]f(x)
and as you said you know the derivatives of both f(x)[sup]2[/sup] and f(x).
Also, the answer is not f’(x) f(x)[sup]3[/sup] = …
What you have written, read in words, is: “the derivative of f times the function f cubed is equal to the derivative of the function f squared.” That this is not what you meant should be clear.
I suggest that you go and talk to your professor about the terminology of calculus. Part of learning calculus is learning the language, and right now you’re not using the language properly. It’s like asking directions to Carnegie Hall by walking up to people and saying “Me going to Hall of Carnegie want to be, I am having you directions ask.” It’s possible to tease out your meaning, but it’s not English, and only someone who knew was very interested and knew what you were trying to say would bother to work it out.
Let’s simplify notation to make the equations clearer.
Let’s substitute y for f(x).
Then the problem is asking to solve d/dx y3
damn stupid message board.
Try again:
Let’s substitute y for f(x).
Then the problem is asking to solve d/dx y[sup]3[/sup].
The hint is that this is equivalent to solving d/dx (y * y[sup2[/sup])
By product rule, this = d/dx y * y[sup]2[/sup] + y * d/dx y[sup]2[/sup]
Now, we do a further product rule application to d/dx y[sup]2[/sup], yielding:
d/dx y[sup]3[/sup] = d/dx y * y[sup]2[/sup] + y * [d/dx y * y + y * d/dx y]
Multiplying through, and combining like terms, we get 3 * y[sup]2[/sup]*d/dx y
Sub back for f(x)
Ok, I think I get it now. I guess all those weird [f(x)^2][f(x)] stuffs made me confused. Now I’ve got to try and find d/dx [f(x)]^4. But I think I see the pattern.
[QUOTE=smiling bandit]
Ok, I think I get it now. I guess all those weird [f(x)^2][f(x)] stuffs made me confused. Now I’ve got to try and find d/dx [f(x)]^4. But I think I see the pattern.
[/QUOTE]
Oh: thanks all! It was big help.
I don’t, however, think I’ll be able to simplify out the answer to the question
Find the derivative of (x^3 -5x^2 +2x -7)^10. 
What am I missing?
If u is a function of x isn’t
**d(u[sup]n[/sup]) = nd(u[sup]n-1[/sup])du
dx dx**
[QUOTE=smiling bandit]
Oh: thanks all! It was big help.
I don’t, however, think I’ll be able to simplify out the answer to the question
Find the derivative of (x^3 -5x^2 +2x -7)^10.
[/QUOTE]
If you have to do that one, it’s worth reading ahead about the chain rule.
[QUOTE=David Simmons]
What am I missing?
If u is a function of x isn’t
**d(u[sup]n[/sup]) = n(u[sup]n-1[/sup])du
dx dx**
[/QUOTE]
Yes (only you included an extra d, which I have taken the liberty of deleting). However, that’s (a special case of) the Chain Rule, which the OP hasn’t gotten to yet.
The OP’s homework may, in fact, be trying to lead him to discover that very pattern—at least in the case where n is a positive integer, in which case your rule could be proved (by mathematical induction) without having to know about the Chain Rule in all its general glory.
[QUOTE=Thudlow Boink]
Yes (only you included an extra d, which I have taken the liberty of deleting). However, that’s (a special case of) the Chain Rule, which the OP hasn’t gotten to yet.
The OP’s homework may, in fact, be trying to lead him to discover that very pattern—at least in the case where n is a positive integer, in which case your rule could be proved (by mathematical induction) without having to know about the Chain Rule in all its general glory.
[/QUOTE]
First of all, I doubt the homework is providing the ground work for the Chain Rule. It actually appears to be providing the induction steps for the Power Rule, using the Produce Rule as its basis.
Whatever it is trying to lay the groundwork for, it would be possible to prove the correctness of the general formula given by ultrafilter through induction without knowing the Chain Rule, if it is possible to prove it by induction at all.
It certainly is possible to use the product rule (and/or induction) and that’s almost certainly what the assignment is supposed to be. Of course, this is homework help which is technically verboten. Regardless, the correct answer is to take a blank sheet of paper and work out the derivative using only the product rule. You want a blank sheet of paper so you don’t try to do anything silly like cram a half dozen lines of work into half a centimeter so it will fit in the space you’ve allotted yourself for this problem…
[QUOTE=Tenebras]
Of course, this is homework help which is technically verboten.
[/QUOTE]
If you read the OP carefully, in this case there is no reason not to provide help because the “homework” involved can be completed through cooperation, etc. (It’s for a college class). 