Partial derivative check me

  • Thread starter rocomath
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  • #1
rocomath
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[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?
 
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Answers and Replies

  • #2
Hootenanny
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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?
I don't think so.

HINT:

[tex]f(x,y) = x^{7/5}y^{4/5}[/tex]
 
  • #3
rocomath
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I don't think so.

HINT:

[tex]f(x,y) = x^{7/5}y^{4/5}[/tex]
omg ... I'm embarassed :D

ok so ...

[tex]f_x(x,y)=\frac 7 5x^{\frac 2 5}y^{\frac 4 5}[/tex]
 
  • #4
Hootenanny
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omg ... I'm embarassed :D

ok so ...

[tex]f_x(x,y)=\frac 7 5x^{\frac 2 5}y^{\frac 4 5}[/tex]
Sounds good to me :approve:
 
  • #5
rocomath
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Thanks Hootenanny :)
 
  • #6
Hootenanny
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  • #7
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You would have gotten the same answer, but the denominator has the wrong exponent.
 

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