What is the partial derivative of f(x,y) with respect to x?

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Homework Help Overview

The discussion revolves around finding the partial derivative of the function f(x,y) = √[5]{x^7y^4} with respect to x. Participants are exploring the correct application of differentiation rules in the context of multivariable calculus.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants present their calculations for the partial derivative and question the correctness of each other's results. There is a hint provided that suggests an alternative expression for the function, prompting a reevaluation of the derivative.

Discussion Status

The discussion includes multiple attempts at calculating the partial derivative, with some participants expressing uncertainty about their results. Hints and alternative forms of the function are shared, indicating a collaborative effort to clarify the correct approach.

Contextual Notes

Some participants express embarrassment over their initial calculations, suggesting a learning process in understanding the differentiation of the given function. There is also mention of a potential error in the exponent of the denominator in one of the responses.

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?
 
Last edited:
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rocomath said:
[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]
 
Hootenanny said:
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]
 
rocomath said:
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:
 
Thanks Hootenanny :)
 
rocomath said:
Thanks Hootenanny :)

A pleasure as always roco :smile:
 
You would have gotten the same answer, but the denominator has the wrong exponent.
 

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