- #1

rocomath

- 1,755

- 1

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

[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:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter rocomath
- Start date

- #1

rocomath

- 1,755

- 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

- 9,624

- 8

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

rocomath

- 1,755

- 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

- 9,624

- 8

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

rocomath

- 1,755

- 1

Thanks Hootenanny :)

- #6

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,624

- 8

Thanks Hootenanny :)

A pleasure as always roco

- #7

Vid

- 402

- 0

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

Share:

- Replies
- 3

- Views
- 404

- Last Post

- Replies
- 5

- Views
- 386

- Replies
- 7

- Views
- 351

- Last Post

- Replies
- 1

- Views
- 526

- Replies
- 5

- Views
- 722

- Replies
- 22

- Views
- 925

- Last Post

- Replies
- 3

- Views
- 341

- Replies
- 1

- Views
- 159

- Replies
- 10

- Views
- 439

- Replies
- 8

- Views
- 408