I refer to Rudin's (Principles of Mathematical analysis, 3rd ed.) definition of differentiability:

Suppose E is an open set in $R^n$ and f maps E into $R^m$ and $x \in E$. If there exists a linear transformation A of $R^n$ into $R^m$ such that $$\lim_{|h| \rightarrow 0} \frac{ |f(x+h) - f(x) -Ah| } {|h|} = 0,$$ then we say that f is differentiable at x and we write $f'(x) = A$.

With this definition and assuming the $l_2$ norm, $f'(x)$ for $(x_1^2 + x_2^2)$ is given by $(2x_1, 2x_2)$. since $$\lim_{|h| \rightarrow 0} \frac{ |f(x+h) - f(x)| } {|h|} = \lim_{|h| \rightarrow 0} \frac{2(x_1h_1 + x_2 h_2)}{|h|} = \lim_{|h| \rightarrow 0} \frac{(2x_1, 2x_2)'(h_1, h_2)}{|h|}.$$

However I could not find $f'(x)$ for $(x_1 + x_2)^2$ with this definition. I get $$\lim_{|h| \rightarrow 0} \frac{ |f(x+h) - f(x)| } {|h|} = \lim_{|h| \rightarrow 0} \frac{2(x_1 + x_2)(h_1 + h_2) + h_1h_2}{|h|}$$

The cross term $h_1 h_2$ becomes a problem, if it were not present, I would write $f'(x) = (2(x_1+x_2), 2(x_1+x_2))$.

I have read elsewhere on this forum that polynomials are differentiable. I would be very thankful I someone were to tell me how do I proceed to find the total derivative, $f'(x)$ for a given polynomial function.

  • $\begingroup$ I'm confused, $(x_{1}^{2}+x_{2}^{2})$ is a two variable polynomial but $f(x)$ seems to denote a one variable function $\endgroup$ – Carlos Laguillo May 9 '15 at 13:39
  • $\begingroup$ @CarlosLaguillo "$x$" can refer to the entire vector $(x_1,x_2)$. Some write $\mathbf x, \hat x, \overrightarrow x$, etc. but it's not imperative as long as you specify what you mean by $x$. $\endgroup$ – GPerez May 9 '15 at 14:02
  • $\begingroup$ I don't think your limit calculations are correct. To start with your final expressions always have $h$, but the limit symbol is gone. Also the limits you calculate are different from the expression you first state. $\endgroup$ – GPerez May 9 '15 at 14:14
  • $\begingroup$ I could not find a mistake in the limit calculations. @GPerez, I have made a few edits, hope that makes things clearer. $\endgroup$ – Sudheer May 11 '15 at 5:22
  • $\begingroup$ Basically I meant what the given answer says, it's not the correct limit. $\endgroup$ – GPerez May 11 '15 at 17:28

Caution: It isn't correct to write $$ f'(x) = \lim_{h \to 0} \frac{\left|f(x + h) - f(x)\right|}{\left|h\right|} $$ even for functions of one variable, and it isn't sensible to write $$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$ for functions of more than one variable, since division by a vector is generally meaningless.

Instead, you must write the condition for multivariable differentiability as $$ 0 = \lim_{h \to 0} \frac{\left|f(x + h) - f(x) - f'(x)(h)\right|}{\left|h\right|}. \tag{1} $$

Separately, the difference quotient computations familiar from one variable aren't as convenient when dealing with functions of several variables. For $f(x_{1}, x_{2}) = x_{1}^{2} + x_{2}^{2}$, you'd have \begin{align*} 0 &= \lim_{h \to 0} \frac{\left|f(x + h) - f(x) - Ah\right|}{\left|h\right|} \\ &= \lim_{h \to 0} \frac{\left|(x_{1}^{2} + 2x_{1}h_{1} + h_{1}^{2} + x_{2}^{2} + 2x_{2}h_{2} + h_{2}^{2}) - (x_{1}^{2} + x_{2}^{2}) - Ah\right|}{\left|h\right|} \\ &= \lim_{h \to 0} \frac{\left|(2x_{1}h_{1} + h_{1}^{2} + 2x_{2}h_{2} + h_{2}^{2}) - Ah\right|}{\left|h\right|}. \end{align*} But what is $Ah$? Well, okay, it has to be the linear terms in $h$; here, $Ah = 2x_{1}h_{1} + 2x_{2}h_{2}$. But technically, you still have to verify that the preceding limit is zero if $A$ is defined as shown.

In practice, it's easier to "guess" the expression $Ah$ by taking partial derivatives, then to verify that (1) holds for the resulting choice of $A$. For the preceding function, we have $$ D_{1}f(x_{1}, x_{2}) = 2x_{1},\qquad D_{2}f(x_{1}, x_{2}) = 2x_{2}, $$ leading us to guess/presume that $Ah = 2x_{1}h_{1} + 2x_{2}h_{2}$. We substitute this in the right-hand side of (1) and check the limit is zero. Recycling the earlier computation, we find $$ \lim_{h \to 0} \frac{\left|f(x + h) - f(x) - Ah\right|}{\left|h\right|} = \dots = \lim_{h \to 0} \frac{\left|h_{1}^{2} + h_{2}^{2}\right|}{\left|h\right|} = \lim_{h \to 0} \left|h\right| = 0. $$ The fact that the limit is zero implies simultaneously that $f$ is differentiable at $x$, and that $f'(x)(h) = Ah$.

Similarly, for the function $f(x_{1}, x_{2}) = (x_{1} + x_{2})^{2}$, you "guess" that $f'(x)(h) = 2(x_{1} + x_{2})(h_{1} + h_{2})$. Substituting into (1) and using your computations gives $$ 0 \leq \lim_{h \to 0} \frac{\left|f(x + h) - f(x) - Ah\right|}{\left|h\right|} = \dots = \lim_{h \to 0} \frac{\left|h_{1}h_{2}\right|}{\left|h\right|} \leq \lim_{h \to 0} \left|h_{2}\right| = 0. $$ (The second inequality comes from the elementary inequality $\left|h_{1}\right|/\left|h\right| \leq 1$ for all $h \neq 0$.)

  • $\begingroup$ thanks for the answer. I missed the elementary inequality. I have used the $l_2$ norm, will things change (for a general problem) if we use a different norm. $\endgroup$ – Sudheer May 11 '15 at 5:30
  • $\begingroup$ You're very welcome. :) I'm also using the $l_{2}$-norm. (Any two norms on a finite-dimensional space determine the same topology/notion of convergence, so for functions of $n$ variables you can use a different norm if you like.) $\endgroup$ – Andrew D. Hwang May 11 '15 at 9:41

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