Q:  Let P(x,y)=x³y⁵
(a) Find the indicated partial derivative with respect to x: P_x(2,1)
(b) Find the indicated mixed partial derivative: P_x,y(1,10)

A:  To find a partial derivative, we establish which letter we are treating like our variable.  In part (a), P_x(2,1) means to take the derivative while treating x as our variable, and then plug in the point (2, 1)…. OK, so treat y like a constant:

P_x(x, y) = 3x²y⁵

Notice how I took the derivative of “x” but the “y” part just came along for the ride like a constant multiplier…

Now plug in (2, 1) for x and y:

P_x(x, y) = 3x²y⁵

P_x(2, 1) = 3(2)²(1)⁵ = 3*4*1 = 12

(b) P_x,y(1,10) is asking us to take the partial derivative with respect to x (which we have done) and then to take the partial derivative with respect to y, and THEN plug in the point (1, 10)…. We know the partial derivative with respect to x is:

P_x(x, y) = 3x²y⁵

Now it is time to take the partial derivative with respect to y… So, treat the x part like a constant multiplier (it won’t change):

P_x,y(x, y) = 3x²(5y⁴) = 15x²y⁴

Now plug in our point:

P_x,y(1,10) = 15(1)²(10)⁴ = 15*1*10000 = 150000