Question

Partial derivative with respect to x: x^4*y^2?

Answer 1

@dumbcow @hartnn @myininaya

Answer 2

Just use chain rule

Answer 3

and product rule

Answer 4

So I got 2x^4*y+4x^3*y^2, is it right?

Answer 5

Product rule apply here
\[(x^4y^2)'=(x^4)'y^2 + (x^4)(y^2)'\]
then, chain rule apply here
\[4x^3 y^2 + x^4*2yy'\]

Answer 6

What about -2x^3*y?

Answer 7

where does it come from?

Answer 8

This is another problem.

Answer 9

You have to do them one by one, don't mix them up.

Answer 10

so, the second problem is \(-2x^3y\) right? take derivative of it, right?

Answer 11

With respect to x, partial derivative again.

Answer 12

Do exactly the same, product rule and then chain rule

Answer 13

"partial"

Answer 14

treat y as constant

Answer 15

oh, partial, OH, I am so sorry,

Answer 16

its like differentating ax^4

Answer 17

yes, hartnn is correct, my stupid mistake. :) follow him, hihihi.. so soooorry

Answer 18

So the partial derivative with respect to x for x^4*y^2 is 4x^3*y^2, right?

Answer 19

yes

Answer 20

And the partial derivative with respect to x for -2x^3*y is -6x^2*y, right?

Answer 21

correct :)

Answer 22

Thank you so much, @hartnn ! Now I know that when it comes to partial derivative, I need to treat the other variable as a constant. Before I didn't know this. Thanks a lot for the big hint.

Answer 23

"every" other variable as constant.
and welcome ^_^