Wikipedia:Reference desk/Archives/Mathematics/2010 November 1

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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)

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)

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)

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)

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)

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)

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)

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).

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)

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)

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)

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)

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

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)

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

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)

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

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

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)

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