Wikipedia:Reference desk/Archives/Mathematics/2007 October 9

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October 9

Three calculus questions

1. I was asked to differentiate $y=x+cot(x)$, so I proceeded as follows: $y'=1+(d/dx)[1/tan(x)]<br>=1+(1/sec^{2}(x))<br>=1+cos^{2}(x)$, but I was told that the answer is $-cot^{2}(x)$. Is there a trigonometric identity I should know but don't? 2. I was asked to differentiate $h(s)=(1/s)-10csc(s)$, so I came up with $h'(s)=-(1/s^{2})-(10/cos(s))$, but I hear it's really $-(1/s^{2})+10csc(s)cot(s)$. Does this hinge on yet another trig identity unknown to me?

3. I was asked to differentiate $(x(x^{2}-1))/x+3$, and I got the answer $(-x^{3}+2x^{2}+7x)/(x+3)^{2}$, but someone told me it's $(2x^{3}+9x^{2}-3)/(x+3)^{2}$. Could someone please tell me if either of us is right, and why?

All help is appreciated. Thank you very much, anon. —Preceding unsigned comment added by 70.23.86.189 (talk) 02:12, 9 October 2007 (UTC)

In the first two, you seem to have made the following mistake:

${\frac {d}{dx}}{\frac {1}{g(x)}}={\frac {1}{{\frac {d}{dx}}g(x)}}$ (with g(x) = tan(x) in the first one, and sin(x) in the second) where what you should be considering is ${\frac {d}{dx}}{\frac {1}{g(x)}}={\frac {d}{dx}}\left[g(x)\right]^{-1}={\frac {d}{dx}}f\left(g(x)\right)$ where f(x) = 1/x, and using the chain rule. In the third, my very quick differentiation gave me an answer different to both the ones you offerred, but closer to the second. The same chain rule trick applies here (or you can use the quotient rule, which is equivalent to the chain and product rules used together), and just be very careful with your algebra. Confusing Manifestation 03:48, 9 October 2007 (UTC)

Actually, for question 3, the second answer offered is correct. In all cases, I think it is easiest to use the quotient rule ${\frac {d}{dx}}\left({\frac {f(x)}{g(x)}}\right)={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}$. For the third question, it will help to write the function as ${\frac {x^{3}-x}{x+3}}$. -- Meni Rosenfeld (talk) 06:30, 9 October 2007 (UTC)

In regards to the first two there are trigonometric identities that you 'should' know. It would be better if you didn't convert to sine and cosine before you differentiated. would probably make your answer look the way it's supposed to since, aside from a missing - sign or two you have the correct answers. The rules for deriving the other four trigonometric functions should be in your text. —Preceding unsigned comment added by 69.54.140.201 (talk) 09:25, 9 October 2007 (UTC)

Actually, his answer for the first question is dead wrong. Don't mislead the OP. –King Bee ^{(τ • γ)} 13:38, 9 October 2007 (UTC)

Strong topology on distribution space

Simple question: Is the strong topology on distribution space the same as the compact open topology? I think so but I am a little out of touch and hence need confirmation. Thank you in advance. twma 03:51, 9 October 2007 (UTC)

The locally convex inductive topology on a test space is not the same as the compact open topology. I am very sure about this. twma 03:40, 10 October 2007 (UTC)

CAS programming on PC

Hi,
Does anyone know a program I can use to create/edit Ti-89 files (.89p) on my computer? Massive thanks for anyone that can tell me!!! --Fir0002 07:59, 9 October 2007 (UTC)

TI-89 series#Programming mentions some, including TIs official software development kit available here. Algebraist 14:21, 9 October 2007 (UTC)

y= form of the general Quadratic polynomial in 2 variables

I figured out how to graph Conic Sections on graphing utility in Cartesian mode. This is how it's done.

$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

Now since we are solving for y

$Cy^{2}+(Bx+E)y+(Ax^{2}+Dx+F)=0$

Note that since we are solving y as two equations of x, we can treat x like a constant in a one variable Quadratic. This means we can apply the Quadratic formula to get our equation. If your not convince try completing the square and isolating y just you would in solving a one-variable quadratic. Be sure to completely simplify.

$y={\frac {-Bx-E\pm {\sqrt {(Bx+E)^{2}-4C(Ax^{2}+Dx+F)}}}{2C}}$

Simplify

$y={\frac {-(Bx+E)\pm {\sqrt {B^{2}x^{2}+2BEx+E^{2}-4ACx^{2}-4CDx-4CF}}}{2C}}$

$y={\frac {-(Bx+E)\pm {\sqrt {(B^{2}-4AC)x^{2}+(2BE-4CD)x+(E^{2}-4CF)}}}{2C}}$

The $\pm$ is why there are two equations, graphically it is obvious why there must be at least two. Are any of you familiar with this? —Preceding unsigned comment added by 69.54.140.201 (talk) 10:04, 9 October 2007 (UTC)

Yes I'm quite familiar with doing things exactly like that.87.102.87.171 13:12, 9 October 2007 (UTC)