# Math Help - derivative proof

1. ## derivative proof

Hi, can anyone help me on this problem?

Suppose that instead of the usual definition of the derivative Df(x) , we define a new kind of derivative, D*f(x), by the forumla

D*f(x) = lim h---> 0 {f^2 (x+h) - f^2 ( x)}/h where f^2 (x) = [f(x)]^2

A) Derive formulas for computing the derivative D* of a sum, difference, product, and quotient. (only one solution is needed, from that I think I can figure out the rest)

B) Express D*f(x) in terms of Df(x)

Thanks

2. Originally Posted by storchfire1X
Hi, can anyone help me on this problem?

Suppose that instead of the usual definition of the derivative Df(x) , we define a new kind of derivative, D*f(x), by the forumla

D^*f(x) = lim h---> 0 {f^2 (x+h) - f^2 ( x)}/h where f^2 (x) = [f(x)]^2

A) Derive formulas for computing the derivative D* of a sum, difference, product, and quotient. (only one solution is needed, from that I think I can figure out the rest)
Here's my attempt at (A), I will only do the sum part.

Consider the differentiable functions $f(x) \mbox { and } g(x)$

$D^*[f(x) + g(x)] = \lim_{h \to 0} \frac {[f(x + h) + g(x + h)]^2 - [f(x) + g(x)]^2}{h}$ .....now i was thinking about using the difference of two squares here, but i couldn't really see the fruits of that labor, maybe someone could do it that way and make it work out to something nicer.

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

$= \lim_{h \to 0} \frac {[f^2(x + h) - f^2 (x) ]+ [g^2 (x + h) - g^2 (x)] + 2[f(x + h)g(x + h) - f(x)g(x)]}{h}$

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

$= D^*f(x) + D^*g(x) + 2 \frac {d}{dx}f(x)g(x)$

B) Express D^*f(x) in terms of Df(x)

Thanks
$D^*f = \lim_{h \to 0} \frac {f^2(x + h) - f^2(x)}{h}$

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

$= \lim_{h \to 0}[f(x + h) + f(x)] \frac {d}{dx} f(x)$

3. Originally Posted by Jhevon
Consider the differentiable functions $f(x) \mbox { and } g(x)$
$D^*f = \lim_{h \to 0} \frac {f^2(x + h) - f^2(x)}{h}$

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

$= \lim_{h \to 0}[f(x + h) + f(x)] \frac {d}{dx} f(x)$
Note that if f is differentiable then then it is continuous.
Thus $= \lim_{h \to 0}[f(x + h) + f(x)]=2f(x)$
and $= \lim_{h \to 0}[f(x + h) + f(x)] \frac {d}{dx} f(x) = 2f(x) \frac {d}{dx} f(x)$

4. Originally Posted by Plato
Note that if f is differentiable then then it is continuous.
Thus $= \lim_{h \to 0}[f(x + h) + f(x)]=2f(x)$
and $= \lim_{h \to 0}[f(x + h) + f(x)] \frac {d}{dx} f(x) = 2f(x) \frac {d}{dx} f(x)$
do you have any suggestions for the first one Plato?

5. Hello, storchfire1X!

Here's the Product Rule . . .

Suppose that instead of the usual definition of the derivative $Df(x)$ ,
we define a new kind of derivative, $D^*\!f(x)$, by the forumla:

$D^*\!f(x) \;= \;\lim_{h\to0}\frac{f^2(x+h) - f^2(x)}{h}$ . where $f^2(x) \,= \,[f(x)]^2$
Let $P(x) \:=\:f(x)\!\cdot\!g(x)$

The numerator of the difference quotient is:
. . $[f(x+h)\!\cdot\!g(x+h)]^2 - [f(x)\!\cdot\!g(x)]^2 \;=\;f^2(x+h)\!\cdot\!g^2(x+h) - f^2(x)\!\cdot\!g(x)$

Subtract and add $f^2(x+h)\!\cdot\!g^2(x):$
. . $f^2(x+h)\!\cdot\!g^2(x+h) - f^2(x+h)\!\cdot\!g^2(x) - f^2(x)\!\cdot\!g^2(x) + f^2(x+h)\!\cdot\!g^2(x)$

Factor:
. . $f^2(x+h)\cdot[g^2(x+h) - g^2(x)] \;+ \;g^2(x)\cdot[f^2(x+h) - f^2(x)]$

Divide by $h$:
. . $f^2(x+h)\!\cdot\!\frac{g^2(x+h) - g^2(x)}{h} \;+\;g^2(x)\!\cdot\!\frac{f^2(x+h) - f^2(x)}{h}$

Take the limit:
. . . . $\lim_{h\to0}\left[f^2(x+h)\!\cdot\!\frac{g^2(x+h) - g^2(x)}{h} \;+\;g^2(x)\!\cdot\!\frac{f^2(x+h) - f^2(x)}{h}\right]$

. . $= \;\lim_{h\to0}\left[f^2(x+h)\right]\cdot\underbrace{\lim_{h\to0}\left[\frac{g^2(x+h) - g^2(x)}{h}\right]}_{\text{This is }D^*g(x)} \;+\;g^2(x)\cdot\underbrace{\lim_{h\to0}\left[\frac{f^2(x+h) - f^2(x)}{h}\right]}_{\text{This is }D^*f(x)}$

Therefore: . $D^*\!P(x) \;=\;f^2(x)\cdot D^*\!g(x) \:+\:D^*\!f(x)\cdot g(x)$

6. Originally Posted by Soroban
Therefore: $D^*\!P(x) \;=\;f^2(x)\cdot D^*\!g(x) \:+\^*\!f(x)\cdot g(x) " alt="D^*\!P(x) \;=\;f^2(x)\cdot D^*\!g(x) \:+\^*\!f(x)\cdot g(x) " />
I am sure that this is a typo.
$D^*\!P(x) \;=\;f^2(x)\cdot D^*\!g(x) \:+\^*\!f(x)\cdot g^2(x) " alt="D^*\!P(x) \;=\;f^2(x)\cdot D^*\!g(x) \:+\^*\!f(x)\cdot g^2(x) " />