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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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    Okay, thanks for your help!2012-05-24

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Is there any advantage/reason to calculate derivatives using the definition of derivative instead of the derivative rules (product, quotient, etc)?

There are situations where the derivative rules do not apply, but the derivative nonetheless exists. For example, $ f(x)= \begin{cases} x^2\sin(1/x) \quad & x\ne 0 \\ 0 & x=0 \end{cases}$ has $f'(0)=0$, although derivative rules do not apply at $0$. Such examples occur with complex numbers as well.

But the main reason students are asked to calculate some derivatives using the definition is to practice the skill of using the definition. The skill is essential for proving things about derivatives of a function when no explicit formula of the function is available.

Apart from such exercises, it is more practical to use the derivative rules.