View Full Version : [SOLVED] Partial derivative ... check me plz
rocomath
Apr23-08, 12:41 PM
f(x,y)=\sqrt[5]{x^7y^4}
f_x(x,y)=\frac 1 5(x^7y^4)^{-\frac{4}{5}}(7x^6y^4)
f_x(x,y)=\frac{7x^6y^4}{5\sqrt[5]{x^7y^4}}
Correct?
Hootenanny
Apr23-08, 12:45 PM
f(x,y)=\sqrt[5]{x^7y^4}
f_x(x,y)=\frac 1 5(x^7y^4)^{-\frac{4}{5}}(7x^6y^4)
f_x(x,y)=\frac{7x^6y^4}{5\sqrt[5]{x^7y^4}}
Correct?
I don't think so.
HINT:
f(x,y) = x^{7/5}y^{4/5}
rocomath
Apr23-08, 12:52 PM
I don't think so.
HINT:
f(x,y) = x^{7/5}y^{4/5}omg ... I'm embarassed :D
ok so ...
f_x(x,y)=\frac 7 5x^{\frac 2 5}y^{\frac 4 5}
Hootenanny
Apr23-08, 01:01 PM
omg ... I'm embarassed :D
ok so ...
f_x(x,y)=\frac 7 5x^{\frac 2 5}y^{\frac 4 5}
Sounds good to me :approve:
rocomath
Apr23-08, 01:02 PM
Thanks Hootenanny :)
Hootenanny
Apr23-08, 01:03 PM
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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