PhilipLethman
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Can Someone Do This problem? I'm confused
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Dr.Peterson
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Did you use the definition of the derivative to find \(\frac{\partial f}{\partial x}(0,0)\)? Please show your work.
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HallsofIvy
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Do you not know what "the definition of the derivative" is?
At (0,y), \(\displaystyle \frac{\partial f}{\partial x}= \lim_{h\to 0} \frac{f(h, y)- f(0,y)}{h}\)
here, we are told that f(x, y), for (x, y) not (0,0), is \(\displaystyle \frac{x^3}{x^2+ y^2}\) sp that \(\displaystyle f(h, y)= \frac{h^3}{h^2+ y^2}\) and f(0, y)= 0 so \(\displaystyle \lim_{h\to 0} \frac{f(h, y)- f(0,y)}{h}= \lim_{h\to 0}\frac{h^3}{(h^2+ y^2)h}\).
What is that limit? That will give you \(\displaystyle \frac{\partial f}{\partial x}(0, y)\) as a function of y. Then do the same thing to find the derivative of that at y= 0.
At (0,y), \(\displaystyle \frac{\partial f}{\partial x}= \lim_{h\to 0} \frac{f(h, y)- f(0,y)}{h}\)
here, we are told that f(x, y), for (x, y) not (0,0), is \(\displaystyle \frac{x^3}{x^2+ y^2}\) sp that \(\displaystyle f(h, y)= \frac{h^3}{h^2+ y^2}\) and f(0, y)= 0 so \(\displaystyle \lim_{h\to 0} \frac{f(h, y)- f(0,y)}{h}= \lim_{h\to 0}\frac{h^3}{(h^2+ y^2)h}\).
What is that limit? That will give you \(\displaystyle \frac{\partial f}{\partial x}(0, y)\) as a function of y. Then do the same thing to find the derivative of that at y= 0.