- #1
dimension10
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I was trying to find the derivative of [itex] \overline{x} [/itex] for some [itex] x \in \mathbb{C} [/itex]
I solved this as
[tex] \frac{\mbox{d}}{\mbox{d}x} \left(\overline{x}\right) = \lim_{h \rightarrow 0}\frac{\overline{x+h}-\overline{x}}{h} [/tex]
[tex] \frac{\mbox{d}}{\mbox{d}x} \left(\overline{x}\right) = \lim_{h \rightarrow 0}\frac{\overline{x+h-x}}{h} [/tex]
[tex] \frac{\mbox{d}}{\mbox{d}x} \left(\overline{x}\right) = \lim_{h \rightarrow 0}\frac{\overline{h}}{h} [/tex]
Now, am I right to say that that we can only simplify this further if we know [itex] \mbox{arg}(\mbox{d}x) [/itex], that is the angle of dx?
Thanks.
I solved this as
[tex] \frac{\mbox{d}}{\mbox{d}x} \left(\overline{x}\right) = \lim_{h \rightarrow 0}\frac{\overline{x+h}-\overline{x}}{h} [/tex]
[tex] \frac{\mbox{d}}{\mbox{d}x} \left(\overline{x}\right) = \lim_{h \rightarrow 0}\frac{\overline{x+h-x}}{h} [/tex]
[tex] \frac{\mbox{d}}{\mbox{d}x} \left(\overline{x}\right) = \lim_{h \rightarrow 0}\frac{\overline{h}}{h} [/tex]
Now, am I right to say that that we can only simplify this further if we know [itex] \mbox{arg}(\mbox{d}x) [/itex], that is the angle of dx?
Thanks.