#1

'ello,

would you kindly derivate the following function:

y= 200x * 0.9^x + 100x^2 * 0.9^x * ln 0.9

thanks in advance.

would you kindly derivate the following function:

y= 200x * 0.9^x + 100x^2 * 0.9^x * ln 0.9

thanks in advance.

#2

no

#3

Do your homework kid.

#4

'ello,

would you kindly derivate the following function:

y= 200x * 0.9^x + 100x^2 * 0.9^x * ln 0.9

thanks in advance.

Do your own homework!

#5

what calculus are u in?

#6

what teh hell is ^?

#7

#8

what teh hell is ^?

to the power of

Times

#9

Hang on I'll try.

Do I need to show work?

Also what variable am I solving for?

Do I need to show work?

Also what variable am I solving for?

#10

that would be nice my'lord Of Donkeys

#11

Dude, do your own homework. It's not a hard question.

#12

Oooh...wait. what do you mean by derivate? I thought it said evaluate.

#13

that's pretty much an "university level" problem.

Is this YOUR homework, or you just try to make a fool from someone here?

Is this YOUR homework, or you just try to make a fool from someone here?

#14

x = 42.

prove me wrong.

prove me wrong.

#15

Oooh...wait. what do you mean by derivate? I thought it said evaluate.

edit nvm

#16

Dude, that's not even hard. I thought it might be somewhat complicated, but then I saw that the ln is just a number, it's not even ln(x). Simple product rules apply.

Let's say you have (3x^2)(4x^5) and you need to derivate it. The rule is:

First multiplied by the derivative of the second, plus second times the derivative of the first. Same rule applies to derivatives with over 2 products.

Let's say you have (3x^2)(4x^5) and you need to derivate it. The rule is:

First multiplied by the derivative of the second, plus second times the derivative of the first. Same rule applies to derivatives with over 2 products.

#17

#18

x = 42.

prove me wrong.

#19

'ello,

would you kindly derivate the following function:

y= 200x * 0.9^x + 100x^2 * 0.9^x * ln 0.9

thanks in advance.

do you mean differentiate?

#20

x = 42.

prove me wrong.

No it doesnt!

Thats the meaning of life =D

#21

It's fairly easy, just check your formulas .

#22

do you mean differentiate?

That's what I took it to mean. Otherwise...

What the **** kind of math are you in?

#23

yeah good old calc 2 have fun Differentiating that nasty son bitch

#24

yeah good old calc 2 have fun Differentiating that nasty son bitch

Calc II? That's material they cover within the first month of Calc I.

#25

No it doesnt!

Thats the meaning of life =D

Therefore in any equation x must equal the answer to life, the universe and everything

#26

y= 200x * 0.9^x + 100x^2 * 0.9^x * ln 0.9

dy/dx = ln(0.9) * 200x * 0.9^x + 200*0.9^x + 100x^2 * (ln(0.9))^2 * 0.9^x + 200x * 0.9^x * ln(0.9)

dy/dx = ln(0.9) * 200x * 0.9^x + 200*0.9^x + 100x^2 * (ln(0.9))^2 * 0.9^x + 200x * 0.9^x * ln(0.9)

#27

That's what I took it to mean. Otherwise...

What the **** kind of math are you in?

i do maths at university thanks.

i'd never heard of/seen the word "derivate" used before, looked it up on wiki, came back the derivate of a function in calculus is it's slope. just making sure.

mmk?

#28

i do maths at university thanks.

i'd never heard of/seen the word "derivate" used before, looked it up on wiki, came back the derivate of a function in calculus is it's slope. just making sure.

mmk?

Well the first derivative is

*dy/dx*and the second derivative is

*d2y/dx^2*. Don't you use those terms? My maths teacher does all the time...

#29

Dive by zero. DUH! Kids these days...

#30

Dude, do your own homework.It's not a hard question.

well excuse me mr-i-took-20-years-of-college-and-major-in-math-so-doing- y= 200x * 0.9^x + 100x^2 * 0.9^x * ln 0.9-problems-are-easy-son

#31

It's as simple as the chain rule

d/dx [f(x)*g(x)] = (f'(x)*g(x) + g'(x)*f(x))

You've got four fcns so think of it like this f'(x) = (f1'(x)*f2(x)*(f3(x)*f4(x))+(f1(x)*f2'(x)*f3(x)*f4(x)) + ... catch my drift?

I'm assuming you were asking how to differentiate the function.

If you're struggling with that... wait till you start doing triple integrals...

d/dx [f(x)*g(x)] = (f'(x)*g(x) + g'(x)*f(x))

You've got four fcns so think of it like this f'(x) = (f1'(x)*f2(x)*(f3(x)*f4(x))+(f1(x)*f2'(x)*f3(x)*f4(x)) + ... catch my drift?

I'm assuming you were asking how to differentiate the function.

If you're struggling with that... wait till you start doing triple integrals...

#32

...I already gave you the answer.

#33

Forgot to add this to my post... you should invest in TI-89 calculator. Saves you a lot of time later.

#34

i do maths at university thanks.

i'd never heard of/seen the word "derivate" used before, looked it up on wiki, came back the derivate of a function in calculus is it's slope. just making sure.

mmk?

Not meaning to insult you. That last part was directed at TS.

Besides, I've taken both Calc AB and BC (not taking a math this year). The problem TS posted is the easiest crap you learn in Calc AB at the beginning of the year.

+1 to the TI-89. It does everything. My friend had one in Calc and I was jealous as hell because he could differentiate and integrate with the press of a few buttons while I had to go around doing crazy U substitutions and burning brain cells.

#35

If you need help with this then I vehemently suggest you drop Calculus.

Only this once will I do this:

y' = -21.0721 * x * (0.9)^x + 200 (0.9)^x - 21.0721x

Only this once will I do this:

y' = -21.0721 * x * (0.9)^x + 200 (0.9)^x - 21.0721x

*Last edited by Muzikh at Apr 27, 2008,*

#36

lol this is the stuff i'm doing in school right now. i've forgotten it all over spring break X_X

#37

would you kindly derivate the following function:

y= 200x * 0.9^x + 100x^2 * 0.9^x * ln 0.9

thanks in advance.

here use this

problem solved

#38

thanks alot guys, especially you darkstar /love