Question

let f(x,y)=(xy^2)/(x^2+y^2), evaluate 0f/0y at the point (1,-2)

Answer 1

So we need to find the partial derivative of f with respect to y?
Is that what the zero's are suppose to be? :)

Answer 2

yes

Answer 3

\[\large f(x,y)=\frac{xy^2}{x^2+y^2}\]

Hmm so it looks like we need to apply the quotient rule.\[\large \frac{\partial f}{\partial y}=\frac{\color{royalblue}{(xy^2)'}(x^2+y^2)-(xy^2)\color{royalblue}{(x^2+y^2)'}}{(x^2+y^2)^2}\]

Understand the setup?
We'll need to take the partial derivatives of the blue terms with respect to y.

Answer 4

Okay how do i do that

Answer 5

Umm ok let's look at the first one.\[\large \frac{\partial}{\partial y}(xy^2)\]When we take a partial, we hold all other variables constant.
So you could think of the problem this way,\[\large \frac{\partial}{\partial y}(cy^2)\]From here we would apply the derivative as normal, and the constant doesn't affect the process.\[\large 2cy\]

But our answer is actually, \(\large 2xy\), the c was just to hopefully prove a point :)

Answer 6

\[\large \frac{\partial f}{\partial y}=\frac{\color{orangered}{(2xy)}(x^2+y^2)-(xy^2)\color{royalblue}{(x^2+y^2)'}}{(x^2+y^2)^2}\]

Can you do the other term using this same idea?
\[\large \frac{\partial}{\partial y}(c^2+y^2)\]

Answer 7

Would it be 2y

Answer 8

So the derivative of the constant became zero?
Yes good job :)

\[\large f_y(x,y)=\frac{\color{orangered}{(2xy)}(x^2+y^2)-(xy^2)\color{orangered}{(2y)}}{(x^2+y^2)^2}\]

Answer 9

is that the final answer?

Answer 10

Don't do any simplification just yet.
From here, plug in your coordinate pair that you were given.

It should be easier to work out if you plug the numbers before simplifying.

So we want this, \(\large f_y(1,-2)\)

So plug in 1's for all your x's,
and -2's for all your y's.

Answer 11

-4/25?

Answer 12

Yay good job! \c:/
That should be our final answer.

Answer 13

Thank you

Answer 14

Could you also show me how to do (x^2+xy+y)^5

Answer 15

With respect to y again?

Answer 16

yes just like the first one

Answer 17

I got -5

Answer 18

Oh same coordinates as the first problem? (1,-2)

Answer 19

(-1,1)

Answer 20

Hmm I'm getting zero.
Lemme double check my work a sec.

Answer 21

what was the partial dx

Answer 22

\[\large (x^2+xy+y)^5\]

Partial with respect to y,\[\large 5(x^2+xy+y)^4(x+1)\]Understand where the (x+1) is coming from?


Partial with respect to x,\[\large 5(x^2+xy+y)^4(2x+y)\]

Answer 23

Okay that is what i got

Answer 24

So for the partial y, simply plugging in x=-1 shows us that it is zero.\[\large 5(x^2+xy+y)^4\color{orangered}{(-1+1)}\]

Answer 25

Oh but for the partial x, yes I'm getting -5 also.
I think that's what you were saying earlier. :)

Answer 26

DO you know how find sketch the level curve f(x,y)=C where f(x,y)=-e^x+y and C=2