# A Freshman Dream Quotient Rule

#### JasMath33

I have been interested in this idea of the FDQR. This idea states the following. I have been trying to see if there is some 2 functions which make this true, but have not found it on research or with trying functions. Does anyone have any insight on this. I think it is just neat and want to know more about what people know about it.

#### DuckAmuck

What ever satisifies:

$$g' = g^2$$

and

$$f' = f \frac{g^2}{g-1}$$

#### micromass

There are infinitely many solutions. Given any function $g$, the associated function can be found by solving the following ODE:

$$f' g (g - g') = (g')^2 f$$

So for example, let $g(x) = e^x$. Then this reduces to $f(x) = 0$.
If $g(x) = x^2$, then
$$f'(x) x^2 (x^2 - 2x) = 4x^2 f(x)$$
or
$$\frac{f'(x)}{f(x)} = \frac{4}{x(x-2)}$$
Integrating yields
$$\log(f(x)) = 2\log(2-x) - 2\log(x)$$
or
$$f(x) = \frac{(2-x)^2}{x^2}$$
In general, given any function $g$ not satisfying $g=g'$ or $g=0$, we can find an associated function $f$ by
$$f(x) = g(x) + e^{\int \frac{g'}{g-g'}}$$

• DuckAmuck

#### Ssnow

Gold Member
$f′g(g−g′)=(g′)^2f$
I think there a minus $f'g(g-g')=-g'^{2}f$ ...

• SammyS

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