Answer:
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
