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

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

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

2. Following is asked:
   become $\alpha ,$ by using $\beta .$ starting with: $$f(x)\left(\frac{1}{f(x)}\right) = {1}$$
3. 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 ) .
4. 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:$\left(\frac{a^r}{b^r}\right) = \left(\frac{a}{b}\right)^r$ $$-2\left(\frac{f'(x)}{f^2(x)}\right)=\left(\frac{1}{f(x)}\right)'$$
   

5. My problem/question

   How can i get rid of the 2?
   We can only use the product rule.

1 Answer

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 $f(x)$.