Asked by Jennifer
I think I have the right answer, but I am not 100% sure how to do the values at (1,-1) and the one below at (0,-1, 1). Please explain how I need to approach how to do these. Thank You!
Calculate the partial derivative @f/@x, @f/@y and
@f/@x | (1,-1), and @f/@y | (1,-1)
f(x,y) = e^(2x+y)
I got f(y) = 0 and f(x) = e^xy
Find @f/@x, @f/@y, @f/@z and their
values at (0,-1, 1)
1.) f(x, y, z) = xy + xz -yz
I got
f(x) = 0
f(y) = 0
f(z) = 0
2.) f(x, y, z) = 2x^0.2 y^0.8 + z^2
Here is what I got.
f(x) = 0.4(y^0.8) / x^0.8
f(y) = 1.6x^0.2 / y^0.2
f(z) = 2z
Calculate the partial derivative @f/@x, @f/@y and
@f/@x | (1,-1), and @f/@y | (1,-1)
f(x,y) = e^(2x+y)
I got f(y) = 0 and f(x) = e^xy
Find @f/@x, @f/@y, @f/@z and their
values at (0,-1, 1)
1.) f(x, y, z) = xy + xz -yz
I got
f(x) = 0
f(y) = 0
f(z) = 0
2.) f(x, y, z) = 2x^0.2 y^0.8 + z^2
Here is what I got.
f(x) = 0.4(y^0.8) / x^0.8
f(y) = 1.6x^0.2 / y^0.2
f(z) = 2z
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