Mathematics desk
< October 31	<< Oct \| November \| Dec >>	November 2 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 1

Trying to solve calculus problems using the Chain Rule

So, I'm not too sure how to do the whole Chain Rule thing using either of the notations (dy/dx or FOG). I'm trying to do stuff myself. Can you help me with this problem? f(x) = (6x-5)⁴. The book says the answer is (24x-5)³. But... why? Thank you. —Duncan^{What I Do} / _{What I Say} 01:21, 1 November 2010 (UTC)[reply]

I assume the problem is to take the derivative? If so, (24x-5)³ is not the correct answer. Are you sure the problem isn't 6(x-5)⁴ and the answer 24(x-5)³?

For the chain rule, you need to think of the function as having an inside and an outside. Working with (19x+1)⁵ as an example, the inside is 19x+1, and the outside is ( )⁵. You start by taking the derivative of the outside: the derivative of ( )⁵ is 5( )⁴. Then you plug the inside back into that: 5(19x+1)⁴. Then you multiply by the derivative of the inside: 19*5(19x+1)⁴.--130.195.2.100 (talk) 02:39, 1 November 2010 (UTC)[reply]

To understand the Chain Rule in relation to the work presented immediately above by 130.195.2.100, see Example II. Dolphin (t) 03:06, 1 November 2010 (UTC)[reply]

According to the chain rule,

{\frac {d}{dx}}(6x-5)^{4}=4(6x-5)^{3}\cdot 6,

But you haven't told us what the question was! Michael Hardy (talk) 04:12, 1 November 2010 (UTC)[reply]

Yes! The question says, take the derivative of y=(6x-5)^4, and the answer is f'(x)=24(6x-5)^3. —Duncan^{What I Do} / _{What I Say} 05:01, 1 November 2010 (UTC)[reply]

You may or may not find the following symbol-pushing useful:

{\frac {dy}{dx}}={\frac {d[(6x-5)^{4}]}{dx}}={\frac {d[z^{4}]}{dx}}\,\,\,({\text{where }}z=6x-5)

={\frac {d[z^{4}]}{dz}}\times {\frac {dz}{dx}}=4z^{3}\times {\frac {d[6x-5]}{dx}}=4z^{3}\times 6=4(6x-5)^{3}\times 6=24(6x-5)^{3}

The "key step" is the one going from the end of the first line to the beginning of the second line. It's the chain rule:

{\frac {d[z^{4}]}{dz}}\times {\frac {dz}{dx}}={\frac {d[z^{4}]}{\cancel {dz}}}\times {\frac {\cancel {dz}}{dx}}={\frac {d[z^{4}]}{dx}}

Informally and very non-rigorously, the chain rule can be stated as "the $dz$ 's cancel". 67.158.43.41 (talk) 07:11, 1 November 2010 (UTC)[reply]

Any introduction to the Chain Rule, using a question like this one, can be made clearer by beginning with a change of variable:

Let $u=(6x-5)$ and therefore ${\frac {du}{dx}}=6$

The question then becomes: If $y=u^{4}$ find ${\frac {dy}{dx}}$

${\frac {dy}{du}}={\frac {d}{du}}(u)^{4}=4(u)^{3}$

The Chain Rule can be expressed as:

${\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}$

${\frac {dy}{dx}}=4(u)^{3}\cdot 6=4(6x-5)^{3}\cdot 6,$

${\frac {d}{dx}}(6x-5)^{4}=24(6x-5)^{3}$ Dolphin (t) 07:25, 1 November 2010 (UTC)[reply]

Or, shortly and painlessly: d(6x-5)⁴=4(6x-5)³d(6x-5)=4(6x-5)³6dx=24(6x-5)³dx. Bo Jacoby (talk) 08:32, 1 November 2010 (UTC).[reply]

Counterexamples & L'Hopital

If I'm not mistaken, proofs of L'Hopital's rule usually rely on the mean value theorem. So suppose we're in a field like Q in which the MVT does not hold. Are there counterexamples to L'Hopital in such cases? Michael Hardy (talk) 04:14, 1 November 2010 (UTC)[reply]

There are. The basic idea is that functions on R with jump discontinuities are still continuous on Q with the standard metric topology if the jumps happen at irrational points. For example define f:Q → R by f(x) = π/n for π/(n+1) < x < π/n and f(0) = 0. This function is continuous on Q. Let g(x) = x. Then

\lim _{x\to 0}f(x)=\lim _{x\to 0}g(x)=0

and f'(x)/g'(x) = 0 for all x, but

\lim _{x\to 0}f(x)/g(x)=1

. Rckrone (talk) 05:50, 1 November 2010 (UTC)[reply]

Looks like that works. Except maybe you'd want f(x) = 1/n for π/(n+1) < x < π/n, with 1 rather than π in the numerator of the value of the function, so that it would be rational-valued. Let's see....that would make the limit 1/π. If we want the limit to be rational as well, then maybe there's more work to do? Michael Hardy (talk) 15:38, 1 November 2010 (UTC)[reply]

You could replace π/n with a rational approximation to it. If you use good enough approximations, the limit will still be 1. Algebraist 17:15, 1 November 2010 (UTC)[reply]

So they'd have to be succesively better approximations as one approaches x. Michael Hardy (talk) 19:50, 1 November 2010 (UTC)[reply]

Did you mean, as x approaches 0? This is of course easy, you can use

f(x)=4\sum _{i=0}^{n}{\frac {(-1)^{i}}{(2i+1)n}}

for π/(n+1) < x < π/n. -- Meni Rosenfeld (talk) 09:30, 3 November 2010 (UTC)[reply]

Yes, that's what I meant. Thank you Rckrone, Algebraist, and Meni Rosenfeld. Michael Hardy (talk) 00:10, 5 November 2010 (UTC)[reply]

Question

What is the statistical likelyhood of humans (Homo sapiens sapiens) having done all unique actions —Preceding unsigned comment added by Deliciousdreams444 (talk • contribs) 16:38, 1 November 2010 (UTC)[reply]

Zero. —Bkell (talk) 18:22, 1 November 2010 (UTC)[reply]

50%: Either they have or they haven't. 84.153.204.6 (talk) 18:28, 1 November 2010 (UTC)[reply]

No human has infrumppeltated a hynosterous yet, despite there being millions of them on the as-yet undiscovered planet Zargastion. 92.29.115.229 (talk) 11:30, 2 November 2010 (UTC)[reply]

The word 'done' is ill-defined, due to special relativity 70.26.152.11 (talk) 23:46, 2 November 2010 (UTC)[reply]