# Math Help - Help with a second partial derivative, please

1. ## Help with a second partial derivative, please

I am doing a chapter on bivariate calculus and so far have done about 30 problems without any mistakes. Yet I've come across one where I can't match the author's answer. Given that I've found a few typos before, I'd like to know if I'm right or wrong, and if wrong, why.

The relevant part of the problem asks:

If z = f(x,y) = x^3 + y^2 - 8xy + (2x^3)(y^2) + 11, find

a) the partial derivative of f with respect to y, and

b) the partial of f [subscript] y (i.e. the 2nd derivative) with respect to y.

The method I am using is to treat all "x"s as constants and to take the derivative of all "y"s.

Thus, when y' means "derivative of y", and when

f(x,y) = x^3 + y^2 - 8xy + (2x^3)(y^2) + 11,

For a) I get

(0) + (y^2)' - (8x)(y)' + (2x^3)(y^2)' + (0)
= 2y - 8x(1) + (2x^3)(2y)
= 2y - 8x + (4x^3)(y),

For b) I use the same approach.

If the partial derivative of f with respect to y

= 2y - 8x + (4x^3)(y),

then

the partial of f [subscript] y with respect to y should be

= 2(y)' - (0) + (4x^3)(y)'
= 2 (1) + (4x^3)(1)
= 2 + 4x^3.

The book, however, gives the answer as 2 + (12x^2)(y), without showing any work.

Am I missing something?

2. Hello, lingyai!

If $z \:= \:f(x,y) \:= \:x^3 + y^2 - 8xy + 2x^3y^2 + 11$, find

a) the partial derivative of $f$ with respect to $y$

b) the partial of $f_y$ with respect to $y$

The method I am using is to treat all x's as constants and to take the derivative of all y's.

For a) I get: $f_y \:= \:0 + (y^2)' - (8x)(y)' + (2x^3)(y^2)' + 0$
. . $= \:2y - 8x(1) + (2x^3)(2y) \:= \:2y - 8x + 4x^3y$ . . . which matches the book's answer.

For b) I use the same approach.

If $f_y \:=\: 2y - 8x + 4x^3y$, then $f_{yy}$ should be:

. . $f_{yy} \:= \:2(y)' - (0) + (4x^3)(y)' \:= \:2 (1) + (4x^3)(1)\:=\:2 + 4x^3$ . . . Correct!

The book, however, gives the answer as: $2 + 12x^2y$ ??

Am I missing something? . . . Not a thing!
The $12x^2y$ looks like they differentiated with respect to x . . . but even that is wrong.

If $f_y\:=\:2y - 8x + 4x^3y$, then: $f_{yx}\:=\:-8 + 12x^2y$

Since $f_x\:=\:3x^2 - 8y + 6x^2y^2$, then $f_{xx} \:=\:6x + 12xy^2$

Their answer doesn't match anything in the problem!
. . It must be just one more typo on their part.

3. Thanks very much for that feedback. Having only started with multivariate calculus today, I was inclined to assume I'd missed something. Not being able to figure it out was troubling.

I'm all in favor of students learning from "spot the deliberate mistake" type answers, but typos in what is supposed to be a self-contained, self-teaching text are very annoying!

Thanks again,

Ken (lingyai)