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

line or a linear equation graph

step-by-step explanation:

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