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# I have a Math exam and I need answers, PLEASE HELP?!!!?

1) Compute all second order partial Derivatives of the functions:

a) f (x,y) = 6 ln (x^2 + y^2)

b) g (x,y) = 9x^2ye^x

The "^" sign means "to the power of".

2) Use the chain rule to find z' (t) at t=2 given that

z = (x+y) ^ -1; x = t^2+1; y=1-t^2

3) Find the critical points of the functions below and classify them as relative minima, relative maxima or saddle points.

a) f (x,y) = (x-1)^2 + y^3 - 3y^2 - 9y + 5

b) g (x,y) = x^3+ y^3 + 3x^2 - 18y^2 + 81y + 5

Please I really need help. I don't understand what the really want and I could really use some serious help... Thanks

### 5 Answers

- 1 decade agoFavorite Answer
For question #1, think about whether or not you need to rewrite the question first. (You do for both!)

For Question #2, the chain rule formula is:

(everything inside)^n-1(everything inside)^1

z'= (x+y)^-2(1+1)

z'= (x+y)^-2(2)

Now plug in the additional info. to complete the problem.

For Question #3, find the derivative and solve for y. Then plug in y to find x. That will give you your critical point(s). (Most of the time, one of those will be a square root and give you a + or - answer, resulting in two points although there is not always two.) Do a line test to verify your points and you will easily see which points are max./min. I don't know and/or have never heard of saddle points, so I can't help you there.

You can also plug the formula into your calculator to get an idea of the max./min. but most likely they will require the line test to get credit.

Hope this helps!

Source(s): I'm in Calculus 2. - norcekriLv 71 decade ago
Serious help? Sure! You've told us where you're stuck, so it's pretty straightforward.

You need to go back to your textbook and look up the mathematical terms you don't understand. I suspect this includes some of

second-order

partial derivative

chain rule

"prime" character (first derivative)

reltive minimum / maximum

saddle point

These last two are the places where the first derivative is zero. A minimum is the lowest point in a concave-up section; a maximum is the highest point in a concave-down section. A saddle point is where the graph levels out at that point, but then goes back to its previous direction (such as a simple cube function, like y=x^3).

With these definitions and your previous work in the class, you should be able to work out the answers.

- 1 decade ago
hello, I'm a high school student and i think i can help a little.(well i'm taking Calc. now)

For # 1 I have no idea what it ask.

But For #2 I have a solution process. (use Chain rule)

First you find derivative of outside; leave inside the same; multiply the deriv. of the inside so...

Rewrite: z=[ (t^2)-(3-t^2) ]^-1 so..

z'= -1[ (t^2)-(3-t^2) ]^-2 x (2t + 2t)

...at this point you may sub.-in for t=2 or simplfy but its best that u substitute in.

ran out of time for #3. sorry i hope i helped.

- mawhinneyLv 44 years ago
Ok to start with you fairly must begin doing larger to your categories. I might say when you've got a sixty five% you might need to ranking within the eighty's or honestly probably within the ninety's to cross. Andif you may have a 70% you might cross via scoring seventy five or eighty.

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- 1 decade ago
Cheating is wrong and I can't help you.

Here are some online places that might help you.