Calculus help

[dchipman]

I have been out of the math game for quite some time and I have a Global Managerial Economics course in the fall for my MBA program. The instructor sent us a few problems he felt we should be able to do in order to pass this course. However, I am stumped.

Here are a couple of the questions:

y=6x^3 - 4x^2 + 5x -12 find dy/dx and d^2y/dx^2

y = ((x^2)/4) - (5x/8) - 5x^-1 - ((3x^-2)/7) + 3x^-3 + 6 find dy/dx and d^2y/dx^2

y= (3x^2 + 4 + 5)/(6x + 3x^-1 + 5)

Any insight on how to do these problems would be great!
Thank you,
Daunielle

 

[Gene]

You need to know that d(a*x^b) = a*b*x^(b-1)

You do that for each term
That makes:
y=6x^3 - 4x^2 + 5x -12
dy/dx = 18x^2 - 8x + 5 and
d^2y/dx^2 = 36x - 8
Study the formula and the results then try it on the rest. You may want to rewrite things like
((x^2)/4) as
(1/4)x^2 to match the formula better

I don't know what you want to do with the last one. I don't see any quick tricks to simplify it (except 4 = 4x?)
If you want dy/dx you need d(a/b) = (b*da - a*db)b^2

 

[dchipman]

Hi again,
4 should be 4x.

when you have a negative exponent, is this how you take the derivitive:

2x^-3 = -6x^-2 (by the way, no laughing at the mathematically impaired)

thanks again
daunielle

 

[Gene]

Not quite. It is still b-1 so d(2x^-3) = -6x^-4dx
'cause -3-1=-4

You still didn't verify that you want the derivitive on the last. if so was the hint enough?

I never laugh. Occasionaly I weep a little :twisted:

 

[dchipman]

hiya glen.

i am trying to find the derivitive of the last problem. i guess that would be helpful when asking you how to do a problem to provide what i am actually looking for.

i gotta tell ya this semester of global managerial economics is not looking good.

however your helpful hints have set off a few reminder bells.

thanks so much
daunielle