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[itex]\frac{\partial V}{\partial s}\frac{ds}{dt}=-\frac{1}{\rho}\frac{dP}{ds}-g\frac{dz}{ds}[/itex] (1.28)

The changes of pressure as a function of time cannot accelerate a fluid particle. This is because the same pressure would be acting at every instant on all sides of the fluid particles. Therefore, the partial differential can be replaced by the total derivative in Eq. (1.28)

[itex]V\frac{dV}{ds}=-\frac{1}{\rho}\frac{dP}{ds}-g\frac{dz}{ds}[/itex]

I can't understand the explanation. Please, help me.