I'm confused. Why do you need to take the derivative? If you're looking for when then why not simply plug in into your equation and solve for ?
What I'm basically trying to do is have transistors turn fully on when a certain resistance is achieved on a dial (potentiometer). Each transistor's middle pin (base pin) has a fixed resistor in series with it, this is to create a voltage drop. As the transistors turn on they allow charge to flow to an led.
The transistors turn on when 5 volts ( this limit is what it's design parameters are, this limit will be referred to as ) is applied to that base pin.
Now for the calculus question:
The formula I am using is the one from the Boards Manual for voltage drop.
It is as follows (I've added in the limit, which is not part of the actual formula:
The only thing that's fixed for that formula in this circuit I am designing is
So my only two variables are and . To find the Derivative, could I just do the quotient rule or do I have to make use of Implicit differentiation? Basically which part of calculus should I be looking into to solve this.
(Just a note, R1 in this case is the dial resistor(potentiometer), R1 in it's self has it's own limit, 0 to 10K (10000 ohms). What I'm searching for is different values for R2 which is an actual fixed resistor when voltage will equal 5volts. I hope you all understand what I'm looking for. I know that what I'm looking for will require calculus, which makes me happy that I can apply it)
put the scheme that u'll try to do (u don't need to derivate anything)
when u have voltage divider ( your two resistors) they are connected in such way that when manipulating with them, you decide how much u wont to put voltage on transistors "base-emitter" connection so that your transistor will no longer be "open switch" and become "closed switch" so your led should glow
it's done by basic laws (don't need to complicate anything)