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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. –  Sasha May 24 '12 at 17:30
    
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 $\frac{\partial u}{\partial x}$? –  Derrick May 24 '12 at 17:36
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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 $\lim_{h \to 0} \frac{ u(0+h,0) -u(0,0)}{h}$. –  Sasha May 24 '12 at 17:40
    
@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 $\frac{\partial u}{\partial x}$? –  Derrick May 24 '12 at 17:48
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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. –  Sasha May 24 '12 at 18:01
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1 Answer

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.

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