# Math Help - Calculate : f'(0)

1. ## Calculate : f'(0)

f is the function :
$
f(x+y)=f(x)+f(y)+xy^{2}+x^{2}y
$

and : ${f}'(0)=1$
Calculate : ${f}'(4)$

2. 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

3. 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

$\chi$ $\sigma$

4. 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

5. 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

$\chi$ $\sigma$

6. 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