# Maximum power transfer derivative

#### Piggins

Joined May 5, 2014
26
I am going through circuit analysis studies and many times there will be derivatives and integrals given but yet I somehow dont seen to be able to solve them myself. Like in that page above it says that the derivative of

p= RL/(RT+RL)^2 * vT^2 is dp/dRL= RT - RL/(RT + RL)^3 * vT^2

I tried to derivate it myself with the chain rule but I got

(RT + RL)^-2 -2RL(RT+RL)^-3 which is not right apparently. Can someone point where I made the mistake?

#### MrAl

Joined Jun 17, 2014
6,485
I am going through circuit analysis studies and many times there will be derivatives and integrals given but yet I somehow dont seen to be able to solve them myself. Like in that page above it says that the derivative of

p= RL/(RT+RL)^2 * vT^2 is dp/dRL= RT - RL/(RT + RL)^3 * vT^2

I tried to derivate it myself with the chain rule but I got

(RT + RL)^-2 -2RL(RT+RL)^-3 which is not right apparently. Can someone point where I made the mistake?
Hi,

Maybe you just need to look up how to take a derivative of a fraction like that.

Note first that VT^2 is a constant, so we can take that outside to start with. What we are left with then is:
RL/(RT+RL)^2

which is a fraction with the variable of interest in both the top and the bottom. So the result will be the derivative of that times the constant VT^2.

The rule for fractions in simple form is:
dy/dx=(Ho*dHi-Hi*dHo)/(Ho*Ho)
where:
Hi is the entire numerator, and
Ho is the entire denominator, and
dHi is the derivative of the entire numerator alone, and
dHo is the derivative of the entire denominator alone.

In this case, we have:
Hi=RL, and
Ho=(RT+RL)^2

You now only need take derivatives of these two and then plug everything into that simple formula for dy/dx to get the result, then put the constant back.

Just for example, Ho*Ho=Ho^2=((RT+RL)^2)^2=(RT+RL)^4

and that will be the denominator of the result.

Of course once you are done with the formula you'll have to simplify the result which takes a little more work. In this particular case the denominator will reduce in power.

In words the formula sounds like this...
"Ho*Ho" is pronounced like someone laughing, ho ho ho ho ho, etc.
"Hi" is pronounced like a common greeting that means Hello.
"dHi" is pronounced "dee high", where "dee" sound like the fourth letter of the English alphabet: "d".
"dHo" is pronounced "dee Ho", where again "Ho" sounds like part of a laugh like Santa Clause.

So the entire formula sounds like this when read out loud:
"Ho dee hi minus hi dee ho, over ho ho".

It's a little comical, so it makes it a little easier to remember that way.

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