Question

Let F(x,y): e^(4xy) * cos(1x) * sin(2y). Compute the following partial derivatives: Fx and Fy

Answer 1

Would Fx be: 4y*e^(4xy)* -sin(x) *sin(2y)?

Answer 2

So you used product rule:
\[F=\sin(2y) \cdot (e^{4y \cdot x}) \cdot (\cos(x)) \\ F_x=\sin(2y)[(e^{4y \cdot x})_x \cdot \cos(x)+ e^{4y \cdot x} (\cos(x))_x]\]

Answer 3

I wrote sin(2y) on the outside of the [ ]
because it was a constant multiple

Answer 4

Now we have \[(e^{4y \cdot x})_x \text{ and } (\cos(x))_x\]

Answer 5

Which it looks like you know from above...

Answer 6

\[F_x=\sin(2y)[4ye^{4yx}\cos(x)+e^{4xy}(-\sin(x))]\]

Answer 7

@tiffany_rhodes are you there?

Answer 8

Yes, sorry I was typing the answer into my hw to check it. It is correct. Thanks!

Answer 9

So basically we can use the product rule for the two functions with x in them and just treat the sin(2y) as a constant because it doesn't have x in the function?

Answer 10

Right

Answer 11

So I used constant multiple rule first just by bringing sin(2y) down
Then I used product rule for the product of the functions of x we had
Then chain rule the e^(4y*x)

Answer 12

I purposely wrote e^(4xy) as e^(4y*x) because it is easy for me to see the constant part in the exponent

Answer 13

Ooh, okay. Thanks again for the help!

Answer 14

Do you think you find F_y?

Answer 15

Yes. I just have to follow the same procedure as Fx but for Fy, just treat the cos(x) as a constant and use the product/chain rule for the functions with y in them?

Answer 16

sounds good

Answer 17

Good luck