Partial derivative by limit definition

Review limit definition

Recall that the partial derivative of f (x, y) with respect to x at the point (a, b) is the same thing as the ordinary derivative of the function g(x) = f (x, b):

$${\frac{{\partial f}}{{\partial x}}}$$

(Here we think of b as just a constant.) We illustrate this graphically as follows.

The green curve can be viewed as the function g(x) and the black line is tangent to the green curve.

You may recall from one-variable calculus how the ordinary derivative was defined. It was with the nice formula

$${\frac{{\mathrm{d} g}}{{\mathrm{d} x}}} \lim_{{h\to 0}}^{} {\frac{{g(a+h)-g(a)}}{{h}}}$$ (1)

You can use the the below CVT to refresh your memory.

Here g(x) is plotted with the thick green curve. The point on the curve with x = a (i.e., the point (a, g(a)) in the (x, y) plane) is plotted as a large black point. The smaller red point shows the point on the curve with x = a + h. The line through (a, g(a)) with slope given by

$${\frac{{g(a+h)-g(a)}}{{h}}}$$ (2)

is shown in blue. (Can you see why this is true? The height of the red ball is g(a + h), the height of the black ball is g(a), so the "rise" is g(a + h) - g(a). The "run" between the points is h. So, rise over run is given by equation (2).)

You can drag the blue point on the slider to decrease h. As h approaches zero, equation (2) approaches the definition (1) of the derivative g'(a). Hence, the slope of the blue line approaches the derivative g'(a). (You can drag the large black point to change a. But you can't drag the smaller red point to change h; you have to use the slider.)

Since

$${\frac{{\partial f}}{{\partial x}}}$$

we can apply the limit defintion of g'(a) to conclude that

$${\frac{{\partial f}}{{\partial x}}} \lim_{{h\rightarrow 0}}^{} {\frac{{f(a+h,b) - f(a,b)}}{{h}}}$$ (3)

In a similar manner, we define

$${\frac{{\partial f}}{{\partial y}}} \lim_{{h\rightarrow 0}}^{} {\frac{{f(a,b+h) - f(a,b)}}{{h}}}$$ (4)

Example with limit definition

Define f (x, y) by

$$\begin{cases} \displaystyle \frac{x^3 +x^4-y^3}{x^2+y^2} & \... ...if } (x,y) \ne (0,0) 0 & \text{if } (x,y) = (0,0) \end{cases}$$

If we want to calculate the partial derivative of f (x, y) at any point away from the origin (0, 0), we can use the usual formulas. However, if we want to calculate $${\frac{{\partial f}}{{\partial x}}}$$(0, 0), we have to use the definition of the partial derivative. (There are no formulas that apply at points around which a function definition is broken up in this way.)

So, we plug in the definition (3). We use the fact that f (0, 0) = 0 and

$${\frac{{h^3+h^4-0^3}}{{h^2+0^2}}} {\frac{{h^3+h^4}}{{h^2}}}$$

Then,

$${\frac{{\partial f}}{{\partial x}}} \lim_{{h\rightarrow 0}}^{} {\frac{{f(0+h,0)-f(0,0)}}{{h}}}$$

$$\lim_{{h\rightarrow 0}}^{} {\frac{{f(h,0)-f(0,0)}}{{h}}}$$

$$\lim_{{h\rightarrow 0}}^{} {\frac{{\displaystyle h+h^2-0}}{{h}}}$$

$$\lim_{{h\rightarrow 0}}^{}$$ = 1.

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