(Px – 20) (100 – 2 Px + Pyg)
Differentiating with respect to P we get
100 – 4Px + Pyg + 40 = 0
I don't understand how this is calc... whenever I multiply through and take the deriv I get another answer. Do you first multiply Px through then 20??
Pyg will be treated as a constant.
It doesn'matter how you multiply it out.
Px*100-2Px^2+PxPyg-2000+40Px+20Pyg=f
100-4Px+Pyg+40=0
How do the 2000 and 20pyg cancel out...the first pyg is still there?
To differentiate the expression (Px – 20) (100 – 2 Px + Pyg) with respect to P, you can use the product rule of differentiation. The product rule states that if you have two functions u(x) and v(x) being multiplied, then the derivative of their product is given by:
(d/dx) (u(x) v(x)) = u'(x) v(x) + u(x) v'(x)
In this case, u(x) = Px – 20 and v(x) = 100 – 2 Px + Pyg. To differentiate the expression, you need to find the derivative of u(x) and v(x) separately, and then apply the product rule.
1. Differentiating u(x) = Px – 20:
We differentiate each term of u(x) separately because the derivative of a constant (like 20) is zero. So, we get:
du/dP = (d/dP) (Px) – (d/dP) (20)
= x – 0
= x
2. Differentiating v(x) = 100 – 2 Px + Pyg:
Here, we need to differentiate each term of v(x) separately:
dv/dP = (d/dP) (100) – (d/dP) (2 Px) + (d/dP) (Pyg)
= 0 – 2x + 0
= -2x
Now, we can apply the product rule:
(d/dP) [(Px – 20) (100 – 2 Px + Pyg)] = (du/dP) v(x) + u(x) (dv/dP)
= (x) (100 – 2 Px + Pyg) + (Px – 20) (-2x)
Simplifying further:
= 100x – 2x^2P + xPyg + 2Px^2 – Pygx – 40x
= xPyg – Pygx + 60x
So, differentiating the expression (Px – 20) (100 – 2 Px + Pyg) with respect to P gives us xPyg – Pygx + 60x.