Expert: Scotto - 1/8/2010

Question
QUESTION: Hi !
i want some help to answer these two questions please !!

1) f(x,y)= xy^2 - ln (xy+1)   , at (0,1)

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


i hope that  you can help me !

ANSWER: The partial derivative means to hold the other variable constant.

1) Treating y as a constant, it can be seen that ∂f/∂x = y² - y/(xy+1)
since the derivative of xy+1 with respect to x is y.

Treating x as a constant, it can be seen that ∂f/∂y = 2xy - x/(xy+1),
since the derivative of xy+1 is x when the derivative is with respect to y.

2) Treating y as a constnat, ∂f/∂x = 2^(3x-5y)•ln(2)•3.  This is the same as 3•ln(2)•2^(3x-5y),
since simpler variables are put first with constants out front.

Treating x as a constant, ∂f/∂y = 2^(3x-5y)•ln(2)•(-5).  This is the same as -5•ln(2)•2^(3x-5y),
since simpler variables .... ya ya ya ... same as last one.


---------- FOLLOW-UP ----------

QUESTION: I'm sorry because i forget to tell you that the question want to find:

fx , fy , fxx , fyy , fxy ?

and the last thing i want to ask you please : where we substitute the points ? after we find fxy ?!!

ANSWER: 1)
fx = ∂f/∂x, and that was given as y² - y/(xy+1).
fy = ∂f/∂y, and that was given as 2xy - x/(xy+1).

fxx = ∂²f/∂x² = y²/(xy+1)².
fyy = ∂²f/∂y² = 2x + x²/(xy+1).
fxy = ∂²f/∂x∂y = -((xy+1) - xy)/(xy+1)² = 2x - 1/(xy+1)².

To find these at (0,1), put in x=0 and y=1 into each equation.

2)
fx = ∂f/∂x, and that was 3*ln(2)*2^(2x-3y).
fy = ∂f/∂y, and that was -5*ln(2)*2^(2x-3y).

fxx = ∂²f/∂x², and that was 3²*ln²(2)*2^(2x-3y)= 9*ln²(2)*2^(2x-3y).
fyy = ∂²f/∂y², and that was (-5)²*ln²(2)*2^(2x-3y) = 25*ln²(2)*2^(2x-3y).
fxy = ∂²f/∂x∂y, and that is -15*ln²(2)*2^(2x-3y).

Here, we put x=1 and y=2 into each equation to find the value of each.


---------- FOLLOW-UP ----------

QUESTION: Can you show me the steps in details with explanation to get these answers????!!!!

Answer
When taking the partials with respect to one of the variables, hold the others as constants.

What is needed to do the above derivatives is the following:
A) If f and g are functions of the variable, d(fg) = fg' + gf'.
B) If f and g are functions of the variable, d[f/g] = (gf' - fg')/g².
C) The derivative of a power z^n is nz^(n-1).
D) The derivative of ln[f(z)] = f'(z)/f(z).
E) The derivative of K^f(x) is ln(K)f'(x)K^f(x).

Using these laws and reviewing back on what I gave, they should be explained.

So to find fxy = ∂²f/∂x∂y. we differentiate with respect to each variable one at a time,
and then get the answer.  When f = 2^(3x-5y) + y^3, fx = ∂f/∂x = 3ln(2)2^(3x-5y).
Since the base is 2, that gives the ln(2). Since there is a 3x in the power, that gives us 3.
Taking this, we can then take the derivative with respect to y to find ∂²f/∂x∂y.
It can be seen that (∂/∂y)(3ln(2)2^(3x-5y)) = ∂²f/∂x∂y = -15*ln²(2)2^(3x-5y).

When differentiating with respect to x and y, it doesn't matter which one is done first.

Basically, just look at the rules and look at the answers until how to do it is seen.