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[Christoff Yurievich]

I know to change the square root to a 1/2 root, and the inside part factors to (4-x)(4+x), but after that it involves product and/or chain rule and I'm still not completely familiar with those rules yet. (if there's an easy way to remember how to do them, I'm all ears).

I need the 1st derivative to find the extrema of the graph. I have that part figured out, I just need the derivative to get it started. Thanks.

oh yeah, I'm working in word with the equation program but for some reason it won't let me do squares (i.e. superscript numbers). Is there a way to make it do this? the keyboard shorcut is ctrl-shift-+ but it doesn't work when I'm making the equations.

 

[GaimeGuy]

if you have two pieces of algebra, a*b = y , y' = (a' * b) + (b' * a).

And remember that when you have algebra inside algebra, you have to do the derivative of the outside algebra, first.

The order is: Power, Trig, Algebra (PTA)

So the derivative of (sin(3x))^5 would be:

5 * (sin(3x))^4 * cos(3x) * 3

in this case, the bolded stuff is the derivative of the power portion of the algebra, sin(3x)^5. The italicized portion is the derivative of the trig portion (in this case, sin, which has a derivative of cos), and then the underlined stuff derivative of the inside algebra of 3x, which equals 3.

I don't know if that was a good explanation or not, but I tried my best.

 

[Citainus]

2(.25)^1 * sqrt(16-x^2) + (.25x^2 * .5(16.x^2)*(16-2x))


It involves product and chain rule.

Product is simple:

xy = x'y +y'x

Chain rule is a bit more:

y(x) = y'(x)* x'

 

[GaimeGuy]

2(.25)^1 * sqrt(16-x^2) + (.25x^2 * .5(16.x^2)*16-2x)


It involves product and chain rule.

Product is simple:

xy = x'y +y'x

Chain rule is a bit more:

y(x) = y(x)' * x'

you mean y'(x) * x'.

 

[Citainus]

you mean y'(x) * x'.

That makes it easier to understand, but I'm not sure if you're supposed to define 'y' as seperated from x, it's still the derivative of the whole function.

 

[GaimeGuy]

That makes it easier to understand, but I'm not sure if you're supposed to define 'y' as seperated from x, it's still the derivative of the whole function.

well, y'(x) is just the standard way I learned it.
And if you have compound equations where they describe stuff like f(g(x)), the way I described makes more sense. I don't see how anything else can make sense.

But eh, whatever works for you.

 

[Christoff Yurievich]

thanks for the help. Now I will apply this to finding the second derivative for the inflection points. I think I can handle it, given the extra tips.

Thanks again. (pretty soon "every damn week" is going to apply to my math threads!)

 

[GaimeGuy]

thanks for the help. Now I will apply this to finding the second derivative for the inflection points. I think I can handle it, given the extra tips.

Thanks again. (pretty soon "every damn week" is going to apply to my math threads!)

just a question: do you live in aussie or something? I don't know how anyplace could be starting a calc class at the moment, midway through the spring semester. I mean, we're preparing for the AP exam over here, which is next week (AP exams = Advance Placement, by the US college board. Basically, AP exams are for high schoolers, and those who pass earn potential college credit)

 

[Aruarian Reflection]

Good luck on the APs! Depending on the eventual college you matriculate into, AP tests could be a big help.

 

[NohWun]

y = .25x^2 * (16-x^2)^1/2

Shouldn't it be:

y' = 2(.25)x * (16-x^2)^1/2 + 0.25x^2 * [ 1/2*(16-x^2)^-1/2 * (0-2x)]

It's been a long time since undergrad... let me know if I'm wrong.

 

[Christoff Yurievich]

just a question: do you live in aussie or something? I don't know how anyplace could be starting a calc class at the moment, midway through the spring semester. I mean, we're preparing for the AP exam over here, which is next week (AP exams = Advance Placement, by the US college board. Basically, AP exams are for high schoolers, and those who pass earn potential college credit)

um, no not really. It's a quarterly college, so the quarter just started about 4 weeks ago. These classes are just to get the last remaining credits for my degree, which I should be getting (hopefully) before 2006.