Math Insight

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 $g(x)=f(x,b)$: \begin{align*} \pdiff{f}{x}(a,b) = g'(a). \end{align*} (Here we think of $b$ as just a constant.) We illustrate this graphically as follows.

Partial derivative as slope

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 \begin{align*} g'(a) = \diff{g}{x}(a) = \lim_{h\to 0} \frac{g(a+h)-g(a)}{h}. \label{eq:limitdef} \end{align*} You can use the the below applet to refresh your memory.

Applet: Ordinary derivative by limit definition

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Applet: Ordinary derivative by limit definition

Ordinary derivative by limit definition. A function $g(x)$ is plotted with a thick green curve. The point $(a,g(a))$ (i.e., the point on the curve with $x=a$) is plotted as a large black point, which you can change with your mouse. The smaller red point shows the point on the curve with $x=a+h$, where you can change $h$ by dragging the blue point on the slider with your mouse. The blue line through the black and red points has slope given by \begin{align} \frac{g(a+h)-g(a)}{h}. \end{align} As you decrease $h$ toward zero, this slope of the blue line approaches the derivative $g'(a)$, as the above expression in this limit is exactly the limit definition of the derivative.

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To understand the formula for the slope of the blue line through the red and black points, note that the height of the red point is $g(a+h)$ and the height of the black point is $g(a)$. Therefore, the “rise” between those points is $g(a+h)-g(a)$. The “run” between the points is $h$. Rise over run is gives the slope of the line \begin{align*} \frac{g(a+h)-g(a)}{h} \end{align*} As $h$ approaches zero, this expression approaches the above definition of the derivative $g'(a)$. Hence, the slope of the blue line approaches the derivative $g'(a)$.

Since \begin{align*} \pdiff{f}{x}(a,b) = g'(a) \end{align*} we can apply the limit defintion of $g'(a)$ to conclude that \begin{align*} \pdiff{f}{x}(a,b) = \lim_{h\rightarrow 0} \frac{f(a+h,b) - f(a,b)}{h}. \label{eq:limitdefx} \end{align*} In a similar manner, we define \begin{align*} \pdiff{f}{y}(a,b) = \lim_{h\rightarrow 0} \frac{f(a,b+h) - f(a,b)}{h}. \label{eq:limitdefy} \end{align*}

Example with limit definition

Define $f(x,y)$ by \begin{align*} f(x,y) = \begin{cases} \displaystyle \frac{x^3 +x^4-y^3}{x^2+y^2} & \text{if } (x,y) \ne (0,0)\\ 0 & \text{if } (x,y) = (0,0) \end{cases} \end{align*} 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 $\displaystyle \pdiff{f}{x}(0,0)$, 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 above limit definition for $\pdiff{f}{x}$. We use the fact that $f(0,0)=0$ and \begin{align*} f(h,0) = \frac{h^3+h^4-0^3}{h^2+0^2} = \frac{h^3+h^4}{h^2} = h+h^2. \end{align*} Then, \begin{align*} \pdiff{f}{x}(0,0) &= \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+h\\ &=1. \end{align*}

This partial derivative is illustrated by a tangent line of slope 1 in the below applet.

Applet: Example partial derivative by limit definintion

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Applet: Example partial derivative by limit definintion

Example partial derivative by limit definintion. The partial derivative of a function $f(x,y)$ at the origin is illustrated by the green line that is tangent to the graph of $f$ in the $x$ direction. The partial derivative $\pdiff{f}{x}(0,0)$ is the slope of the green line. The partial derivative at $(0,0)$ must be computed using the limit definition because $f$ is defined in a piecewise fashion around the origin: $f(x,y)= (x^3 +x^4-y^3)/(x^2+y^2)$ except that $f(0,0)=0$. You can change the point $(x,y)$ at which $\pdiff{f}{x}(x,y)$ is evaluated by dragging the red point. Since $\pdiff{f}{x}(x,y)$ is discontinuous around $(0,0)$, the derivative jumps as soon as you move the point. Press the Home key to return to $(0,0)$.

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