Proof of P = Fv.

I would like to add remarks/comments pointwise on your comments which are as follows:
I begin with the formula for kinetic energy: $E_k=\frac12​mv^2$ 

Taking the derivative with respect to velocity, I find $\frac{dE_k}{dv}=mv$ , which gives the rate of change of kinetic energy with respect to velocity. This expression is correct.
I interpret $mv$ as the change in kinetic energy per unit increase in velocity. This statement is incorrect, as it doesn't represent change in kinetic energy per unit increase in velocity. It is change in kinetic energy per unit change in velocity.
To connect this to power, I add the following reasoning:Power is energy transferred per second. If we assume a 1 m/s increase in velocity happens in 1 second,  then mv represents the amount of energy added per second in this scenario. This assumption corresponds to an acceleration of $1 m/s^2$. For a different acceleration, we can scale the result proportionally. For example, if the acceleration is $a$, the energy change per second scales to $amv$, as the rate of velocity increase is now $a$ rather than $1$. This interpretation in the form of example as well as generalised result is correct and the same can be shown mathematically as follows. $$\frac{dE_k}{dv}=mv$$ Multiplying and dividing by $dt$ in the denominator we can rewrite it as: $$\frac{dE_k}{dt\cdot\frac{dv}{dt}}=mv$$ $$\frac{dE_k}{dt\cdot a}=mv$$
A driving force is applied to increase the object's velocity over time. Simultaneously, a resistive force (like  friction) opposes the motion. These forces alternate, balancing out to maintain constant velocity. By imagining the alternation of driving and resistive forces happening over smaller and smaller time intervals $x\to 0$, I treat the kinetic energy as a boundary condition that fluctuates around a stable value. The power delivered by the driving force can then still be understood in terms of kinetic energy changes. In this setup, $P = Fv$ emerges naturally when scaling up or down based on acceleration or force. This approach connects power and kinetic energy more explicitly than conventional methods, which often skip this link. The power formula no doubt ultimately comes after the above step $$\frac{dE_k}{dt}=(ma)v$$ $$P=Fv$$ The issue is that it is longer derivation (mathematically 2 step longer derivation than the standard derivation which takes the derivative both sides with respect to time instead of velocity which you did) What you did is like $2+2 = 1+3=4$

To me this seems far more logical than traditional explanations, but I am conernced that I have gone wrong somewhere. If this is a sound explaination then why isn't it used (I can't find it anywhere) as it seems to connect kinetic energy to power far better than traditional methods. This is not used anywhere because there doesn't exist any physical quantity which is defined as change in energy per unit velocity. It is not normal for any individual to understand what that thing could be. And it is not correct as well, because energy can be changed even without change in velocity but it is always true that when energy is transferred/changed there has to be change in time absolutely. This is the key I hope you get the idea. Scientists have discovered change in energy per unit time is Power. Even though it is a different interpretation of yours which might be convenient for you to think like that but it is of no use in practical life.

You were almost correct everywhere but ultimately you arrive at the correct result but in longer and the harder way.