1. ## Calculate : f'(0)

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

2. Originally Posted by SADANE ANTAR
f is the function :
$\displaystyle f(x+y)=f(x)+f(y)+xy^{2}+x^{2}y$
and : $\displaystyle {f}'(0)=1$
Calculate :$\displaystyle {f}'(4)$
If the derivative exists at $\displaystyle x$ then:

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

CB

3. Originally Posted by CaptainBlack
If the derivative exists at $\displaystyle x$ then:

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

CB
According to ...

Derivative -- from Wolfram MathWorld

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

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

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

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

4. Originally Posted by chisigma
According to ...

Derivative -- from Wolfram MathWorld

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

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

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

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

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...

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

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

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

6. Originally Posted by chisigma
If we accept the 'more symmetrical' notation of the derivative...

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

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

Merry Christmas from Italy

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

CB