Question

If f(x, y) = 49 − 7x2 − y2 , find fx(1, 2) and fy(1, 2) and interpret these numbers as slopes. fx(1, 2) = fy(1, 2) =

Answer #1

please comment if you have any doubts will clarify

part 1)
Find the partial derivatives of the function
f(x,y)=xsin(7x^6y):
fx(x,y)=
fy(x,y)=
part 2)
Find the partial derivatives of the function
f(x,y)=x^6y^6/x^2+y^2
fx(x,y)=
fy(x,y)=
part 3)
Find all first- and second-order partial derivatives of the
function f(x,y)=2x^2y^2−2x^2+5y
fx(x,y)=
fy(x,y)=
fxx(x,y)=
fxy(x,y)=
fyy(x,y)=
part 4)
Find all first- and second-order partial derivatives of the
function f(x,y)=9ye^(3x)
fx(x,y)=
fy(x,y)=
fxx(x,y)=
fxy(x,y)=
fyy(x,y)=
part 5)
For the function given below, find the numbers (x,y) such that
fx(x,y)=0 and fy(x,y)=0
f(x,y)=6x^2+23y^2+23xy+4x−2
Answer: x= and...

Find all values of x and y such that
fx(x, y) = 0 and fy(x, y) = 0
simultaneously.
f(x, y) = x2 + 3xy + y2 − 19x − 16y +
40
(x,y) = ( ___ , ___ )

(1 point)
Find all the first and second order partial derivatives of
f(x,y)=7sin(2x+y)−2cos(x−y)
A. ∂f∂x=fx=∂f∂x=fx=
B. ∂f∂y=fy=∂f∂y=fy=
C. ∂2f∂x2=fxx=∂2f∂x2=fxx=
D. ∂2f∂y2=fyy=∂2f∂y2=fyy=
E. ∂2f∂x∂y=fyx=∂2f∂x∂y=fyx=
F. ∂2f∂y∂x=fxy=∂2f∂y∂x=fxy=

Consider the function f(x,y) = xe^((x^2)-(y^2))
(a) Find f(1,−1), fx(1,−1), fy(1,−1). Use these values to find a
linear approximation for f (1.1, −0.9).
(b) Find fxx(1, −1), fxy(1, −1), fyy(1, −1). Use these values to
find a quadratic approximation for f(1.1,−0.9).

Let f(x, y) = 2x^3y^2 + 3xy^3 4x^3 y. Find
(a) fx
(c) fxx
(b) fy
(d) fyy
(e) fxy
(f) fyx

You are given that the function f(x,y)=8x2+y2+2x2y+3 has first
partials fx(x,y)=16x+4xy and fy(x,y)=2y+2x2, and has second
partials fxx(x,y)=16+4y, fxy(x,y)=4x and fyy(x,y)=2. Consider the
point (0,0). Which one of the following statements is true?
A. (0,0) is not a critical point of f(x,y).
B. f(x,y) has a saddle point at (0,0).
C. f(x,y) has a local maximum at (0,0).
D. f(x,y) has a local minimum at (0,0).
E. The second derivative test provides no information about the
behaviour of f(x,y) at...

f(x, y) =
ln
x +
x2 + y2
; fx(5,
−3)

Question about using the convolution of distribution:
1. we have the formula: integral fx(x)fy(z-x)dx=integral
fx(z-x)fy(x)dx
I know this are equivalent. However, how do I decide which side
I should use ?
For example,X~Exp(1) and Y~Unif [0,1] X and Y independnt and the
textbook use fx(z-x)fy(x)dx.
However, can I use the left hand side fx(x)fy(z-x)dx???is there
any constraint for using left or right or actually both can lead me
to the right answer???
2. For X and Y are independent and...

4.4-JG1 Given the following joint density function in Example
4.4-1:
fx,y(x,y)=(2/15)d(x-x1)d(y-y1)+(3/15)d(x-x2)d(y-y1)+(1/15)d(x-x2)d(y-y2)+(4/15)d(x-x1)d(y-y3)
a) Determine fx(x|y=y1) Ans: 0.4d(x-x1)+0.6d(x-x2)
b) Determine fx(x|y=y2) Ans: 1d(x-x2)
c) Determine fy(y|x=x1) Ans: (1/3)d(y-y1)+(2/3)d(y-y3)
d) Determine fx(y|x=x2) Ans:
(3/9)d(y-y1)+(1/9)d(y-y2)+(5/9)d(y-y3)
4.4-JG2
Given fx,y(x,y)=2(1-xy) for 0 a) fx(x|y=0.5) (Point Conditioning)
Ans: (4/3)(1-x/2)
b) fx(x|0.5

Let joint CDF Fx,y (x,y) = сxy(x2 + y2)
for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.
Find а) constant с.
b) Fx|y (x|y) for x = 0.5, y = 0.5.

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