Take the partial differential equations of 3x^4y^2 - e^(xy) what would f_xyy be? I got this wrong and It's pissing me off. Any help would be appreciated.
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Hi,
If f(x,y) = 3 x^4 y^2 -e^(xy),
then to get f_xyy, we do the following:
First partial differentiate f with respect to x (treating y as a constant) to get
12 x^3 y^2 -ye^(xy) = f_x.
Then partial differentiate f_x with respect to y treating x as a constant to get
24 x^3 y - xy e^(xy) -e^(xy) = f_xy.
Finally, partial differentiate f_xy one more time with respect to y, treating x as a constant, to get
24 x^3 -x^2 y e^(xy) - x e^(xy) -x e^(x y).
When partial differentiating the term -xy e^(xy), we use the product rule, which is
d/dy(-xy e^(x y)) = -[d/dy(xy e^(x y))] (here we just pull out the -1, which we can do because the partial differential operator is a linear operator, and so we can pull out any constants as we did with -1)
Using the product rule,
d/dy(xy e^(x y)) = xy d/dy e^(xy) + e^(xy) d/dy (xy)
=x y x e^(xy) + e^(xy)x
which after simplification, is equal to
x^2 y e^(x y) + x e^(xy),
and so,
-d/dy(xy e^(x y)) = -[x^2 y e^(x y) + x e^(xy) ]
= -x^2 y e^(x y) - x e^(xy) after distributing the negative one in this last expression.
A useful fact for product rule, which may be useful to you in the near future, is if you need to different a product of say, three functions,
v(x) = g(x)h(x)z(x), v'(x) is just equal to
g'(x)h(x)z(x) + g(x)h'(x)z(x) + g(x)h(x)z'(x)
all you do is right the expression "g(x)h(x)z(x)" three times (because there are three functions) separated by '+' signs, then put a prime on each function once as was done above. This rule generalizes to a product of n functions. If you want to different a product of n functions, right the product n times (with each term separated by a plus sign), then put a prime on the first function in the first term, then go the second term and put a prime on the second function in the second term, then go and put a prime on the third function in the third term, ... then go and put a prime on the (n-1) th function in the (n-1)th term, and finally put a prime on the nth function in the nth term, and walla, your differentiation of the product of n functions is complete!
I hope this helps you out!
If f(x,y) = 3 x^4 y^2 -e^(xy),
then to get f_xyy, we do the following:
First partial differentiate f with respect to x (treating y as a constant) to get
12 x^3 y^2 -ye^(xy) = f_x.
Then partial differentiate f_x with respect to y treating x as a constant to get
24 x^3 y - xy e^(xy) -e^(xy) = f_xy.
Finally, partial differentiate f_xy one more time with respect to y, treating x as a constant, to get
24 x^3 -x^2 y e^(xy) - x e^(xy) -x e^(x y).
When partial differentiating the term -xy e^(xy), we use the product rule, which is
d/dy(-xy e^(x y)) = -[d/dy(xy e^(x y))] (here we just pull out the -1, which we can do because the partial differential operator is a linear operator, and so we can pull out any constants as we did with -1)
Using the product rule,
d/dy(xy e^(x y)) = xy d/dy e^(xy) + e^(xy) d/dy (xy)
=x y x e^(xy) + e^(xy)x
which after simplification, is equal to
x^2 y e^(x y) + x e^(xy),
and so,
-d/dy(xy e^(x y)) = -[x^2 y e^(x y) + x e^(xy) ]
= -x^2 y e^(x y) - x e^(xy) after distributing the negative one in this last expression.
A useful fact for product rule, which may be useful to you in the near future, is if you need to different a product of say, three functions,
v(x) = g(x)h(x)z(x), v'(x) is just equal to
g'(x)h(x)z(x) + g(x)h'(x)z(x) + g(x)h(x)z'(x)
all you do is right the expression "g(x)h(x)z(x)" three times (because there are three functions) separated by '+' signs, then put a prime on each function once as was done above. This rule generalizes to a product of n functions. If you want to different a product of n functions, right the product n times (with each term separated by a plus sign), then put a prime on the first function in the first term, then go the second term and put a prime on the second function in the second term, then go and put a prime on the third function in the third term, ... then go and put a prime on the (n-1) th function in the (n-1)th term, and finally put a prime on the nth function in the nth term, and walla, your differentiation of the product of n functions is complete!
I hope this helps you out!