Determine the derivative of the reciprocal function, only with the product rule. 
*

*Following is given:

$( \alpha )$ the derivative of a inverse function: $$\left(\frac{1}{f}\right)' = -\frac{f'}{f^2}$$

$( \beta )$ The the product rule: $$({f}*g) = ({f'}*{g})+({f}*{g'})$$


*Following is asked:

become $\alpha ,$ by using $\beta .$ starting with: $$f(x)\left(\frac{1}{f(x)}\right) = {1}$$

*General information about the function
If the function f(x) is derivable in the open interval ]a,b[ and f(x) isn't zero in ]a,b[, then is the inverse function $\left(\frac{1}{f}\right)$ derivable in ]a,b[ then applies ( \alpha ) . 

*My solution on (2) 
 starting with: $f(x)\left(\frac{1}{f(x)}\right) = {1}$ 

First: apply the product rule we become:
$$f'(x)\left(\frac{1}{f(x)}\right) + f(x)\left(-\frac{1}{f^2(x)}\right)  = {0}$$ 

that's the same as: $f'(x)\left(\frac{1}{f(x)}\right) - f(x)\left(\frac{1}{f^2(x)}\right)  = {0}$ 

Second; apply the product rule again:
$$f''(x)\left(\frac{1}{f(x)}\right)+f'(x)\left(-\frac{1}{f^2(x)}\right)-f'(x)\left(\frac{1}{f^2(x)}\right)+f(x)\left(-\frac{2*1}{f^3(x)}\right)=0$$

that's the same as:
$f''(x)\left(\frac{1}{f(x)}\right)-f'(x)\left(\frac{1}{f^2(x)}\right)-f'(x)\left(\frac{1}{f^2(x)}\right)-f(x)\left(\frac{2*1}{f^3(x)}\right)=0$

Third; simplify the result: (put equal terms first)
$$\left(\frac{f'(x)}{f(x)}\right)[1-\left(\frac{1}{f(x)}\right)]-\left(\frac{1}{f^2(x)}\right)[f'(x) - \left(2\frac{f(x)}{f(x)}\right)]$$

that's the same as:
$\left(\frac{f'(x)}{f(x)}\right)[1-\left(\frac{1}{f(x)}\right)]-\left(\frac{1}{f^2(x)}\right)[f'(x) - \left(\frac{2}{1}\right)]$

More simplification ( \beta ): 
$$\left(\frac{f'(x)}{f(x)}\right)-\left(\frac{f'(x)}{f^2(x)}\right)-\left(\frac{f'(x)}{f^2(x)}\right) + \left(\frac{2}{f^2(x)}\right)=0$$

$$\left(\frac{f'(x)}{f(x)}\right)-2\left(\frac{f'(x)}{f^2(x)}\right)+\left(\frac{2}{f^2(x)}\right)=0$$

$$\left(\frac{f'(x)}{f(x)}\right)-2\left(\frac{f'(x)}{f^2(x)}\right) 
[1-\left(\frac{1}{f'(x)}\right)] =0$$

Last; the solution: 
  Take the First and third part of the equation. Bring it to the right part.
  We get: 
$$-2\left(\frac{f'(x)}{f^2(x)}\right) 
[1] =\left(\frac{f'(x)}{f'(x)f(x)}\right)$$
That's is the same as: (make it cleaner and use the following identities).
 The identity:$\left(\frac{a^r}{b^r}\right) = \left(\frac{a}{b}\right)^r$
$$-2\left(\frac{f'(x)}{f^2(x)}\right)=\left(\frac{1}{f(x)}\right)'$$


*My problem/question 
How can i get rid of the 2? We can only use the product rule.


At the start there is an error, namely:
$$f'(x)\left(\frac{1}{f(x)}\right) + f(x)\left(-\frac{1}{f^2(x)}\right)  = {0}$$ 
Instead just write, using the product rule with $g(x)= 1/f(x)$: 
$$f'(x)\left(\frac{1}{f(x)}\right) + f(x)\left(\frac{1}{f(x)}\right)'  = {0}$$ 
Now  you have an equation involving $\left(\frac{1}{f(x)}\right)'$.
Just transform the equation to express $\left(\frac{1}{f(x)}\right)'$ in terms of $f'(x)$ and $f(x)$.
