[SOLVED] Chain rule.

**el123** · October 26th 2009, 08:09 PM

$f(x) = \frac{2}{3} \sqrt{36-p^2)}$

$f'(x) =\frac{1}{2}(36-p^2)^{-\frac{1}{2}} \cdot -2p$

Am i doing this right? Im a bit lost.

**VonNemo19** · October 26th 2009, 08:11 PM

Originally Posted by el123

$f(x) = \frac{2}{3} \sqrt{36-p^2)}$

$f'(x) =\frac{1}{2}(36-p^2)^{-\frac{1}{2}} \cdot -2p$

Am i doing this right? Im a bit lost.

Here you have a function of x. This implies that p is either

a) A function of x
or
b) a constant

I think you should reread your instructions.

**PatrickFoster** · October 26th 2009, 08:11 PM

What happened to the 2/3 ?

**ANDS!** · October 26th 2009, 08:12 PM

Where did you get the leading 1/2 from?

**el123** · October 26th 2009, 08:14 PM

Well isn't square root to the power of half? I thought i had to factor the 2/3 out cos its a constant.>??

**VonNemo19** · October 26th 2009, 08:15 PM

Originally Posted by el123

Well isn't square root to the power of half? I thought i had to factor the 2/3 out cos its a constant.>??

Before you worry about that, read my last post.

**ANDS!** · October 26th 2009, 08:22 PM

Lol.

Anyway, he or she is asking a valid point (although it really doesn't have much an impact at this stage of game): your original function is in terms of x, but the variable you are differentiating is "p". Is it actually F(p)?

As for your process, square root is the power of 1/2 - what is being asked, is where did the 2/3 go? You don't need to factor is, just multiply it by 1/2 (and then simplify some more with the -2p).

**PatrickFoster** · October 26th 2009, 08:22 PM

Ok 2 Things:

First, VonNemo19 said, if you are taking the derivative with respect to x for a function that does not have an x in it, then the whole thing is zero (since it can be treated like a constant). But then again, it depends on your instructions.

Second, assuming it's actually f'(p) and no f'(x), then you are correct to bring down the 1/2, but you are incorrect to remove the 2/3 since it is in the same term (via multiplication). So the 1/2 and 2/3 should cancel for the twos, and you should end up with 1/3.

The correct answer would be:

$<br /> -\frac{2}{3}x(36-x^2)^{-\frac{1}{2}}<br />$

**el123** · October 26th 2009, 08:25 PM

My apologies , I hadn't understood what he was getting at , make that function in terms of p.

Thank you ANDS! , now it makes sense.

**VonNemo19** · October 26th 2009, 08:25 PM

Originally Posted by ANDS!

Lol.

(although it really doesn't have much an impact at this stage of game):

It is EXTREMELY relevant at this stage in the game.

**el123** · October 26th 2009, 08:28 PM

Yes , well , very sorry to be rude. Frustration will do that to yah.

**ANDS!** · October 26th 2009, 08:29 PM

Originally Posted by VonNemo19

It is EXTREMELY relevant at this stage in the game.

Agree to disagree. He or she has more than enough time to learn proper Math-English (assuming they did make a typo - which I'm pretty sure they did, as I never saw any problem in Calc 1 that would have been sneaky like that). But I'm not interested in getting into a pissing match here.

**PatrickFoster** · October 26th 2009, 08:33 PM

Calc 1 has implicit differentiation, correct? Or was that in Calculus 2? Regardless, if this were an implicit differentiation problem, the answer could be signficantly different if there is more information in the instructions.

On the other hand, an error was made in the differentiation no matter what the instructions were (no instructions would cause the 2/3 to disappear like that), and so obviously that is something which should be addressed.

**VonNemo19** · October 26th 2009, 08:35 PM

Originally Posted by ANDS!

Agree to disagree. He or she has more than enough time to learn proper Math-English (assuming they did make a typo - which I'm pretty sure they did, as I never saw any problem in Calc 1 that would have been sneaky like that). But I'm not interested in getting into a pissing match here.

Listen guys. I don't know why everyone here is demoniing me right now, nor do I care, but the fact remains

Is p a function of x, or is it constant? Once the piece of business is taken care of, I'll differentiate this bad boy and buy everybody a drink.

Peace.

**el123** · October 26th 2009, 08:37 PM

To be clear, i had made a typo. And i misunderstood what vonnemo has posted , so my reply was to what i had taken as someone taking the piss. Not an overreaction.

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