Question

derivative of 2xye^-x^2

Answer 1

you have two variable expression
so are you doing partial derivatives?

Answer 2

if so are you doing the partial w.r.t. x or y?

Answer 3

and in your drawing why is there a dy at the end?

Answer 4

are we doing the partial w.r.t to y?

Answer 5

like is this what you meant?
\[\frac{\partial }{ \partial y}(2xye^{-x^2})\]

Answer 6

yes respect to y

Answer 7

yes yes @myininaya

Answer 8

ok well that means you treat everything by y like a constant

Answer 9

so in other words you know \[\frac{d}{dy}cy=c \frac{d}{dy}=c \frac{dy}{dy}=c(1)=c\]

Answer 10

im confused with this "treat y as a constant" you mean? take out y from the equation?

Answer 11

\[2xe^{-x^2} \frac{\partial }{\partial y}y=?\]

Answer 12

no treat everything BUT y like a constant

Answer 13

that is the only other variable is x
treat x like a constant

Answer 14

so you see there for your partial i pulled out the constant multiple

Answer 15

all you have to do is find the derivative of y w.r.t to y now

Answer 16

x's the only variable..

Answer 17

can i use the method udv+vdu?

Answer 18

I don't understand. your expression is in terms of x and y.
You have two variables.
You are finding the partial w.r.t y so that means x is like a constant.
so 2x is like a constant here
so e^(-x^2) is also like a constant here
and the product of constants is a constant
so 2xe^(-x^2) is like a constant here

Answer 19

\[\frac{\partial }{ \partial y} (2xye^{-x^2})= 2xe^{-x^2} \frac{\partial }{\partial y} y =\text{ you can finish this }\]

Answer 20

my answer is 2xe^-x^2??