Without $f$ being specified, can we obtain $\frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{x+h}-\frac{1}{x}}{h}$? "Show that $\frac{f(x+h)-f(x)}{h} = -\frac{1}{x(x+h)}$." is the problem I am looking at. In the solutions, the equality $\frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{x+h}-\frac{1}{x}}{h}$ is given as the first step. How does this work? $f$ is not specified in the question. This question is given in the context of introducing differential calculus. Thank you in advance for any help!
 A: The manipulation done only depends on $h \neq 0$, without it requiring any special conditions on $f\left(x\right)$, including that it even be differentiable, even though only certain functions of $f\left(x\right)$ will work, as indicated in the comments and the answer by H Huang. I suspect what I show below was the intent of the solution step.
To see how to make the change without knowing $f\left(x\right)$, treat $x$ as a variable and $h$ as a constant, and note the partial fraction decomposition of the right side of what you're trying to show may be written as
$$\cfrac{-1}{x\left(x + h\right)} = \cfrac{A}{x} + \cfrac{B}{x + h} = \cfrac{Ax + Ah + Bx}{x\left(x + h\right)} \tag{1}\label{eq1}$$
Putting the $2$ terms with $x$ together, then comparing coefficients of $x$ and the constant term, we thus have that
$$A + B = 0 \tag{2}\label{eq2}$$
$$Ah = -1 \tag{3}\label{eq3}$$
From \eqref{eq3}, we get that $A = \frac{-1}{h}$, so \eqref{eq2} then gives $B = \frac{1}{h}$. Substituting these values into \eqref{eq1} gives
$$-\cfrac{1}{x\left(x + h\right)} = \cfrac{-1}{hx} + \cfrac{1}{h\left(x + h\right)} = \cfrac{\frac{1}{x + h} - \frac{1}{x}}{h} \tag{4}\label{eq4}$$
This shows that the right hand sides of the $2$ equations in the question for $\frac{f\left(x + h\right) - f\left(x\right)}{h}$ are equivalent to each other.
Note it's simpler & easier to see this relationship going the other way, by having both terms use a common denominator, simplifying and dividing by the common $h$ factor to get
$$\cfrac{1}{h}\left(\cfrac{1}{x + h} - \cfrac{1}{x}\right) = \cfrac{1}{h}\left(\cfrac{x - \left(x + h\right)}{x\left(x + h\right)}\right) = \cfrac{1}{h}\left(\cfrac{-h}{x\left(x + h\right)}\right) = -\cfrac{1}{x\left(x + h\right)}  \tag{5}\label{eq5}$$
A: We know that the definition of the derivative of $f(x)$ is $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$. Given that $\frac{f(x+h)-f(x)}{h} = -\frac{1}{x(x+h)}$, that means that $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}-\frac{1}{x(x+h)}$. The limit $\lim_{h \to 0}-\frac{1}{x(x+h)}$ is pretty easy to evaluate; it's equal to $-\frac{1}{x^2}$. 
Thus, $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=-\frac{1}{x^2}$.
$\frac{1}{x^2}$ happens to be the derivative of $\frac{1}{x}$, so reversing the 
definition of the derivative, $f(x)$ is $\frac{1}{x}$. 
Thus, even without $f(x)$ being given, from the fact that $\frac{f(x+h)-f(x)}{h}=-\frac{1}{x(x+h)}$, we can find $f(x)$. Knowing $f(x)$, it's not hard to see where $\frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{x+h}-\frac{1}{x}}{h}$ comes from.
