Question

Please help me figure out how to take the derivative of "Y" in respect to "P_Y" of this equation. I've been trying to figure it out for a while, but I can't figure it out.

Also, I do know that the answer is provided here. While I know the answer I'm not quite following how they did what they did. I think I understand how they got the first part of the answer. However, I'm totally lost on how they got the second part (the (-2)(3P_Y) part). Please help me!

3P

Answer 1

A simple chain rule of differentiation is used

As Y* = (I-2Py)/(3Py)

So Y* = (I-2Py)*(3Py)^-1

using chain rule, first differentiate (3Py)^-1 keeping (I-2Py)

& Then differentiate (I-2Py) keeping (3Py)^-1

so dY*/dPy = {(I-2Py)*d(3Py)^-1/dPy} + {(3Py)^-1*d(I-2Py)/dPy}

= (I-2Py)*-1*(3Py)^-2*3  + (3Py)^-1*(-2)

= -3*(3Py)^-2*(I-2Py) - 2*(3Py)^-1