If f (x, y) = x (x2 + y2) - 3/2 esin (x2y), find fx (1, 0). [Hint: Instead of finding fx (x, y) first, note that it's easier to use the following equations.] fx (a, b) = g' (a) where g (x) = f (x, b) fx (a, b) = lim h→0 f (a + h, b) - f (a, b) h

fx (1,0) = 3

Step-by-step explanation:

for the function

f (x, y) = x (x2 + y2) - 3/2 e^[sin (x2y) ]

then defining the function g (x) = f (x, b) we have that fx (x, b) = g' (x) and thus fx (a, b) = g' (a), therefore by definition of the derivative

fx (a, b) = lim h→0 (g (a+h) - g (a)) / h = lim h→0 (f (a+h, b) - f (a+h, b)) / h

since sin (0) = 0, we have that

fx (a, b) = lim h→0 (f (a+h, b) - f (a+h, b)) / h

lim h→0 ((1+h) * [ (1+h) ² - 0²) - 3/2*e^0) - ((1) * [ (1) ² - 0²) - 3/2*e^0) / h =

lim h→0 [ ((1+h) ³ - 3/2) - (1-3/2) ] / h = lim h→0 [ (1+h) ³-1]/h =

lim h→0 (h³ + 3h²*1 + 3h*1² + 1 - 1) / h) =

lim h→0 ((h³ + 3h²+3h) / h) = lim h→0 (h² + 3h+3) = 0 + 3*0 + 3 = 3

thus fx (1,0) = 3