- #1

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**[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?

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- Thread starter rocomath
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- #1

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[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

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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

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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

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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

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Thanks Hootenanny :)

- #6

Hootenanny

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Thanks Hootenanny :)

A pleasure as always roco

- #7

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

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