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

= 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 54.  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,
 * $$ \frac{d}{dx} (6x - 5)^4 = 4(6x - 5)^3 \cdot 6, $$
 * $$ \frac{d}{dx} (6x - 5)^4 = 4(6x - 5)^3 \cdot 6, $$

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)