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

Relevance

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.
• 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

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.

• 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.

• 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.