Calculate : f'(0)

**SADANE ANTAR** · December 13th 2009, 10:14 PM

f is the function :
$ f(x+y)=f(x)+f(y)+xy^{2}+x^{2}y $
and : ${f}'(0)=1$
Calculate : ${f}'(4)$

**CaptainBlack** · December 14th 2009, 04:42 AM

Originally Posted by SADANE ANTAR

f is the function :
$ f(x+y)=f(x)+f(y)+xy^{2}+x^{2}y $
and : ${f}'(0)=1$
Calculate : ${f}'(4)$

If the derivative exists at $x$ then:

$ f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x-h)}{2h}=$ $\lim_{h \to 0}\frac{f(x)+f(h)+xh^2+x^2h-f(x)-f(-h)+xh^2+x^2h}{2h}=$ $ \lim_{h \to 0}\frac{f(h)-f(-h)+2x^2h}{2h}=f'(0)+x^2$

CB

**chisigma** · December 14th 2009, 04:59 AM

Originally Posted by CaptainBlack

If the derivative exists at $x$ then:

$ f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x-h)}{h}=$ ...

CB

According to ...

Derivative -- from Wolfram MathWorld

... the 'more symmetrical' notation of the derivative is...

$f^{'} (x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$

... in any case congratulations for Your elegant procedure!

...

Merry Christmas from Italy

$\chi$ $\sigma$

**CaptainBlack** · December 14th 2009, 06:45 AM

Originally Posted by chisigma

According to ...

Derivative -- from Wolfram MathWorld

... the 'more symmetrical' notation of the derivative is...

$f^{'} (x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$

... in any case congratulations for Your elegant procedure!

...

Merry Christmas from Italy

$\chi$ $\sigma$

Oppss ... ( a hangover from a earlier attempt with a non-symmetric form) It should be correct now

CB

**chisigma** · December 14th 2009, 06:59 AM

If we accept the 'more symmetrical' notation of the derivative...

$f^{'} (x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x-h)}{2h}$ (1)

...do we also have to accept that for a function like $f(x)=|x|$ is $f^{'} (0)=0$ ?... or not?

...

Merry Christmas from Italy

$\chi$ $\sigma$

**CaptainBlack** · December 14th 2009, 01:08 PM

Originally Posted by chisigma

If we accept the 'more symmetrical' notation of the derivative...

$f^{'} (x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x-h)}{2h}$ (1)

...do we also have to accept that for a function like $f(x)=|x|$ is $f^{'} (0)=0$ ?... or not?

...

Merry Christmas from Italy

$\chi$ $\sigma$

$f'(0)=1$ implies that the derivative exists at $0$ , which excludes the posibility of something like $|x|$ . Also this can be done with the derivative from the right being $1$ at $x=0$ (as implied in my comment about an earlier version using an asymmetric form for the difference quotient).

CB

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