Calculus Students: Need Help with Precalc Lab
 

Right now I am working on a precalc lab, and part of the lab asks to find a calculus student to find the derivative of a function that I wrote for one of the problems. I have no idea what a derivative even is, so... any calculus students out there willing to help?

The function is:
y = inverse tan(20x/ (4800 + x^2))

...and just so you know, I wrote "inverse tan" because I couldn't write "tan -1" in superscript.

Disclaimer: If I find out that the above function is incorrect, I may need someone to do it all over again for me...hopefully that won't happen!

-- Jaine

Re: Calculus Students: Need Help with Precalc Lab
 

y' = (-20*(x^2 - 4800))/(x^4+10000*x^2+23040000)

Re: Calculus Students: Need Help with Precalc Lab
 

The derivative of the function, f, above is

f' = [-20(x^2-4800)]/[x^4+10000*x^2+23040000]

However, the derivatives of arc functions are generally not taught in precalculus. It's a calculus subject, so I don't see why your teacher would assign such a problem. I don't think anyone could explain what derivatives are to you (much less go through all the steps to derive the arctan derivative using logarithmic differentiation) without taking many days or weeks.

[edit] Whoops, while I was away, someone already posted the answer... [/edit]

Re: Calculus Students: Need Help with Precalc Lab
 

No don't worry --
My teacher didn't assign us this problem... he asked us to "find a calculus student and have them find the derivative of this function". He wasn't asking us to do something we didn't know how to do; more likely he was trying to bug some of his calc students by having us follow them around asking for derivatives. :)

Thank you both soo much though!

-- Jaine

Re: Calculus Students: Need Help with Precalc Lab
 

ask the ti-89....it normally gives me good answers....

Re: Calculus Students: Need Help with Precalc Lab
 

Haha...I can relate to that. Sure, I know how to do derivitives myself, but punching 2nd 8 is just that much easier.

Re: Calculus Students: Need Help with Precalc Lab
 

Heh, yeah....that's how I got my answer above so quickly. Doing the derivative of the arctangent and then applying the chain rule is quite annoying IMHO.

Re: Calculus Students: Need Help with Precalc Lab
 

awwwww no fair guys! i'm over here stuck w/ an 83!!! meh :rolleyes:
i mean, i was gonna check ur work and stuff BY HAND but now i wont bother cuz you used ur CALCULATOR
...cheaters... ;)
~Stephanie

Re: Calculus Students: Need Help with Precalc Lab
 

no, we just use the right tools...it's like using a bandsaw instead of a hack saw...

Re: Calculus Students: Need Help with Precalc Lab
 

All i have to say is Maple....Maple 9 that is...

If you don't know what maple is, its cause you math is still too easy.

Re: Calculus Students: Need Help with Precalc Lab
 

I hate Maple...

I had to get up for my only class of the day every wednesday at 8 am just so I could go waste an hour using it.

Re: Calculus Students: Need Help with Precalc Lab
 

Just for basic derivative finding heres what u do:
Take the power and multiply it by the coefficient, then lower the power by one. Its that easy. However it does get more difficult once you get into the qutient, product, and chain rules; as well as when u start combining trig (like u did) and wit hparametric equations. good luck

Re: Calculus Students: Need Help with Precalc Lab
 

you don't need luck

all you need is a good AP test to scare you

and the trig is easy:

function derivative
sin cos
cos -sin
tan sec^2
cot -csc^2
sec sec * tan
csc -csc * cot

anyone see the patern? :cool:

and on the calculator issue:
but doing it by hand is the fun way, when you get really good you can laugh at all the mortals who take forever to do the problem w/ a calculator

*evil laugh*

that's all

Re: Calculus Students: Need Help with Precalc Lab
 

The "co" functions are all negative. Also, Maple is for sissies, MatLab is a real man's tool. :)

A derivative is the rate of change of the function. Or more precisely, if f' is the derivative of f, and f is differentiable at a (more on that some other time) f'(a) is the slope of the tangent line of the function where x = a.

Also, Yan, I don't think you need logarithmic differentiation to solve this (you are thinking of integrals maybe?) only implicit differentiation

proof:
to find dy/dx of the function y=atan(x)

first move the tan to the other side
tan(y) = x

take the derivitive of both sides
sec^2(y) * dy/dx = dx/dy

devide over sec^2(y) so that
dy/dx = 1/sec^2(y) * dx/dy

remember trig?
since sec^2(y) = 1 + tan^2(y)
dy/dx = 1/(1 + tan^2(y)) * dx/dy

and tan(y) = x (from the top)
dy/dx = 1/(1 + x^2) * dx/dy

math is fun

-Andy