Work differential: Why $\mathrm{d}W = f \mathrm{d}s$ not $\mathrm{d}W = f \mathrm{d}s + s\space \mathrm{d}f$ Starting from the formula for work given a constant force $W = f s$, if you take the differential of both sides you would expect to get:
$$\mathrm{d}W = f \mathrm{d}s + s \space \mathrm{d}f$$
by an application of the product rule on the right hand side when taking the differential.
However, clearly this is incorrect, as by intuition the force across a segment $ds$ remains approximately constant, leading to the correct answer:
$$\mathrm{d}W = f \mathrm{d}s$$
Why did directly taking the differential on both sides not work? I suspect it might be related to how $f$ is itself a function of $s$.
 A: You can't just take the case of constant force to define work. Work is a form of energy, which is defined ( in one dimension )  by integrating Newton's second law with respect to the spatial coordinate.
$$F=ma = m \frac{dv}{dt}= m \frac{dv}{ds} \frac{ds}{dt} =mv \frac{dv}{ds}$$
So
$$\int_{s_0}^{s_f} F ds =\int_{s_0}^{s_f} mv \frac{dv}{ds}=\int_{v_0}^{v_f} mv dv  = \frac m2 (v_f^2-v_i^2)$$
It is the left side of this equation that should be taken as the definition of work
$$W \equiv \int_{s_0}^{s_f} F ds $$
A: The complete differential is actually the correct definition of Work. The usage usually depends on the what kind of force is exerted. For example, in a system with impulse force, you will account for the change in force as well.
Additionally, the formula changes depending on whether the final energy stored is potential energy or kinetic energy.
A: As you have mentioned, a constant force, thus the second term on the right side expression is redundant, don't you think? The product rule is correct though.
