If $f(x)$ and $g(x)$ are both differentiable and $f(x)g(x) = 1$ for ALL $x$, show $\left[\frac{f’(x)}{f(x)} + \frac{g’(x)}{g(x)}\right] = 0$ I am able to conjure up a simple example such as $f(x) = x$, and $g(x) = \frac1x$, which will compute to $0$, but I am asking for help to write a more rigorous proof, not more rigorous examples, but fluency that that says why it works. Perhaps share any vital Theorems that would apply and how.
 A: $fg=1$ hence $f'g+fg'=(fg)'=0$ so $(f'g+fg')/(fg)=0$ thus $f'/f+g'/g=0$.
A: $h(x)=\log(f(x))+\log(g(x))=\log(f(x)g(x))=\log(1)=0$. It's differentiable due to differentiability of $f(x),g(x)$. Differentiate it on both sides, you will get the result.
A: Consider function $h(x)=\ln(f(x)g(x))$. It's a constant function, so $h'(x)=0$ for all $x \in \mathbb{R}$. But on the other hand if you calculate derivative you get:
$$h'(x)=(\ln(f(x)g(x)))'=(\ln(f(x))+\ln(g(x)))'=\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}.$$
A: If $f(x)$ and $g(x)$ are both differentiable and $f(x)g(x) = 1$, then we have  $0 = 1' = (f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$, showing that $f'(x)g(x) + f(x)g'(x) = 0$.  $f(x)g(x) = 1$ implies $f(x) \ne 0 \ne g(x)$, so we meaningfully have $g(x) = (f(x))^{-1}$ and $f(x) = (g(x))^{-1}$.  Inserting these into $f'(x)g(x) + f(x)g'(x)$ yields the desired result $f'(x)(f(x))^{-1} + g'(x)(g(x))^{-1} = 0$.
Nothing fancy.  It's merely the product rule (mostly).
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!,
A: Yeah, the (non-linear) differential operator $f\mapsto \frac{f'}{f}$ ( the logarithmic derivative) takes products to sums. It's the differential analogue of:
$a \%$ increase followed by $b \% $ increase is (approximately) $(a+b)\%$ increase.
