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May 19th, 2010, 02:18 AM   #1
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Partial derivatives

Hi,

I have a question:

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Let f(x,y) = x^2/x+y

Obtain the first two partial derivatives f_x and f_y and evaluate them at point (2,2).

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Could someone please confirm my answer?

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f_x; make y constant.
= 2x(x + y)^-1 + x^2(-x + y)^-2

f_x = 2x / (x + y) + x^2 / (-x + y)^2
If we sub (2,2) we have:

2(2) / (2 + 2) + 2^2 / (-2 + 2)^2 = 1 + 0 = 1 ANS

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f_y; make x constant.
= 0(x + y)^-1 + x^2(-x + y)^-2

f_y = x^2 / (-x + y)^2
If we sub (2,2) we have:

2(2) / (-2 + 2)^2 = 0 ANS

--

.. ?

wulfgarpro.
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May 19th, 2010, 02:30 AM   #2
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Re: Partial derivatives

With y constant d/dx x^2(x + y)^-1 = 2x(x + y)^-1 - x^2(x + y)^-2

With x constant d/dx x^2(x + y)^-1 = -x^2(x + y)^-2
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May 20th, 2010, 02:54 AM   #3
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Re: Partial derivatives

Evaluated at point (2,2), I get:

f_x = 12/16 = 3/4

f_y = -4/16 = -1/4

?

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May 20th, 2010, 02:58 AM   #4
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Re: Partial derivatives

Correct.
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