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Let $f(x+iy)=x^2-y^2 + 5xi$. So hence $u(x,y)=x^2-y^2$ and $v(x,y)=5x$

In my notes it calculated $\frac{\partial u}{\partial x}$ at $0$ as follows:

$\frac{\partial u}{\partial x}(0,0)=\displaystyle\lim_{h\rightarrow 0}\frac{u(x+h,y)-u(x,y)}{h} \\=\displaystyle\lim_{h\rightarrow 0}\frac{u(h,0)-u(0,0)}{h}\\=\displaystyle\lim_{h\rightarrow 0}\frac{h^2}{h}=\displaystyle\lim_{h\rightarrow 0}h=0$

But is it possible to calculate $\frac{\partial u}{\partial x}$ at $0$ by just finding that $\frac{\partial u}{\partial x} = 2x$, and then substituting $x=0,y=0$ and thus getting $\frac{\partial u}{\partial x}=0$?

If so, it seems easier that way rather than taking limits as above.

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    Yes, it is possible to use the result $\frac{\partial u}{\partial x}$ and do the substitution. The results will be the same since $u$ is differentiable.2012-05-24
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    Thanks @Sasha , but is there any advantage or reason as to why one would use $\frac{\partial u}{\partial x}(0,0)=\displaystyle\lim_{h\rightarrow 0}\frac{u(x+h,y)-u(x,y)}{h}$ instead of just calculting $\frac{\partial u}{\partial x}$?2012-05-24
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    The limit you wrote is the definition of $\frac{\partial u}{\partial x}(x,y)$, not of $\frac{\partial u}{\partial x}(0,0)$. The definition of the latter is $\lim_{h \to 0} \frac{ u(0+h,0) -u(0,0)}{h}$.2012-05-24
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    @Sasha , Whoops sorry, my bad. I meant, is there any advantage/reason to use $\frac{\partial u}{\partial x}(x,y)=\displaystyle\lim_{h\rightarrow 0}\frac{u(x+h,y)-u(x,y)}{h}$ instead of calculating $\frac{\partial u}{\partial x}$?2012-05-24
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    If you can obtain $\frac{\partial u}{\partial x}$ algebraically, it is a preferred way, otherwise, the definition in terms of the limit may be used to work out the result from the first principles.2012-05-24
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    Okay, thanks for your help!2012-05-24

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