Derivative of formula for battery's output power is not right

Question

There is something about output power of battery in circuits that bothering me. I hope you help me. So we know the formula for power in electrical circuits is (current x voltage) or: $$ P=IV $$ and the voltage of a battery while there is a current running through it is: $$ V = \varepsilon - Ir $$ where $\varepsilon$ is the electromotive force, and $r$ is the internal resistance of the battery. Now if we replace $V$ in the power equation with the voltage of battery, we have: $$ P = I(\varepsilon - Ir) = I\varepsilon-rI^2 $$ if we make a function out of it like this: $$ P(I) = I\varepsilon - rI^2 $$ and graph it, we get something like this for example:

My problem is the derivative of this function: $$ \frac{dP}{dI} = \varepsilon - 2rI $$ which is clearly not equal to the output voltage of the battery, but as far as I know, it should be. If there is something I'm mistaking about, please correct me

Answer 1

Take a look at the equaiton for power: $$P=IV.$$ It may seem at first that one can take the derivative of this expression with respect to the current and obtain: $${dP(I)\over dI}=V,$$ however, this overlooks the fact that $V$ is a function $V=V(I)$, e.g. as you not in your own post: $$V=\epsilon-Ir.$$ So that in reality you have no reason to expect that the derivative of power with respect to current is $V$, rather: $${dP(I)\over dI}={dI\over dI}V+I{dV\over dI}=(\epsilon-Ir)+I(-r)=\epsilon-2Ir.$$ Where I have used the "product rule", $${dfg\over dx}={df\over dx}g+f{dg\over dx},$$ for taking the derivative of a product of functions. You don't have to use the product rule, you can indeed take the derivative by writing out the function for $P$ explicitly then differentiating, like you did in your example; however, the product rule can be handy for more complicated examples. Thus, your work was correct except for your expectation that the derivative of $P$ should be $V$.