- #1

- 1,752

- 1

**[SOLVED] Partial derivative ... check me plz**

[tex]f(x,y)=\sqrt[5]{x^7y^4}[/tex]

[tex]f_x(x,y)=\frac 1 5(x^7y^4)^{-\frac{4}{5}}(7x^6y^4)[/tex]

[tex]f_x(x,y)=\frac{7x^6y^4}{5\sqrt[5]{x^7y^4}}[/tex]

Correct?

Last edited:

- Thread starter rocomath
- Start date

- #1

- 1,752

- 1

[tex]f(x,y)=\sqrt[5]{x^7y^4}[/tex]

[tex]f_x(x,y)=\frac 1 5(x^7y^4)^{-\frac{4}{5}}(7x^6y^4)[/tex]

[tex]f_x(x,y)=\frac{7x^6y^4}{5\sqrt[5]{x^7y^4}}[/tex]

Correct?

Last edited:

- #2

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,621

- 6

I don't think so.[tex]f(x,y)=\sqrt[5]{x^7y^4}[/tex]

[tex]f_x(x,y)=\frac 1 5(x^7y^4)^{-\frac{4}{5}}(7x^6y^4)[/tex]

[tex]f_x(x,y)=\frac{7x^6y^4}{5\sqrt[5]{x^7y^4}}[/tex]

Correct?

HINT:

[tex]f(x,y) = x^{7/5}y^{4/5}[/tex]

- #3

- 1,752

- 1

omg ... I'm embarassed :DI don't think so.

HINT:

[tex]f(x,y) = x^{7/5}y^{4/5}[/tex]

ok so ...

[tex]f_x(x,y)=\frac 7 5x^{\frac 2 5}y^{\frac 4 5}[/tex]

- #4

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,621

- 6

Sounds good to meomg ... I'm embarassed :D

ok so ...

[tex]f_x(x,y)=\frac 7 5x^{\frac 2 5}y^{\frac 4 5}[/tex]

- #5

- 1,752

- 1

Thanks Hootenanny :)

- #6

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,621

- 6

A pleasure as always rocoThanks Hootenanny :)

- #7

- 401

- 0

You would have gotten the same answer, but the denominator has the wrong exponent.

- Last Post

- Replies
- 0

- Views
- 862

- Last Post

- Replies
- 1

- Views
- 544

- Replies
- 2

- Views
- 1K

- Replies
- 1

- Views
- 639

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 819

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 7

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 0

- Views
- 2K