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I am currently learning about partial derivatives, and while I was watching a lecture noticed something interesting. The example question was to find the values of:
$$f(x,y) = \frac{x+y}{x^2+y^2}$$ at the position $(2, 0)$. I found that by substituting the value of y into the formula which made differentiation much quicker/easier and then subbing x in, I got the same answer for the x component of the partial derivative as if I was to assume y to be fixed then differentiate the question through the quotient rule, then sub in the values.

Method 1:

$$f(x,0)=\frac{1}{x}$$ $$f_x(x,0)=-\frac{1}{x^2}$$ $$f_x(2,0)=-\frac{1}{4}$$

Method 2: $$f(x,y)=\frac{x+y}{x^2+y^2}$$ $$f_x(x,y)=\frac{-x^2-2yx+y^2}{\left(x^2+y^2\right)^2}$$ $$f_x(2,0)=-\frac{1}{4}$$

My question is, is the method that I used valid or just a a random occurrence?

ArchieMcKay
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    it's true in general, and you should try to write out the definition of partial derivative at a point as a limit to see why the two methods give the same answer. – peek-a-boo Oct 03 '19 at 06:47

1 Answers1

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Assuming $y$ to be fixed is exactly the same thing as substituting a particular value for $y$, so yes, this is a valid method.

Hans Lundmark
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