Given function f(x,y) = e^{(−x²−y²)}

Step 1 :

First partial derivatives : f_x(x,y) and f_y(x,y)

f_x(x,y) = (∂/∂x)f(x,y)

f_x(x,y) = (∂/∂x)e^{(−x²−y²)}

= e^{(−x²−y²)}(∂/∂x)(−x²−y²)

= e^{(−x²−y²)}(−2x−0)

= −2xe^{(−x²−y²)}

f_y(x,y) = (∂/∂y)f(x,y)

f_y(x,y) = (∂/∂y)e^{(−x²−y²)}

= e^{(−x²−y²)}(∂/∂y)(−x²−y²)

= e^{(−x²−y²)}(−0−2y)

= −2ye^{(−x²−y²)}

Step 2 :

Evaluation of the first partial derivatives : f_x(0,0) and f_y(0,0)

f_x(x,y) = −2xe^{(−x²−y²)}

Substitute x=0 , y=0

f_x(0,0) = −2(0)e^{(−0²−0²)}

f_x(0,0) = 0

f_y(x,y) = −2ye^{(−x²−y²)}

Substitute x=0 , y=0

f_y(0,0) = −2(0)e^{(−0²−0²)}

f_y(0,0) = 0

Step 3 :

The second partial derivatives : f_xx(x,y),f_yy(x,y) andf_xy(x,y)

f_xx(x,y) = (∂/∂x)f_x(x,y)

f_xx(x,y) = (∂/∂x)(−2xe^{(−x²−y²)})

= ((-2x)e^{(−x²−y²)}(−2x−0))+(−2e^{(−x²−y²)})

= ((4x²)e^{(−x²−y²)})+(−2e^{(−x²−y²)})

= ((4x²-2)e^{(−x²−y²)})

= 2(2x²-1)e^{(−x²−y²)}

f_xy(x,y) = (∂/∂y)f_x(x,y)

f_xy(x,y) = (∂/∂y)(−2xe^{(−x²−y²)})

= ((-2x)e^{(−x²−y²)}(−0−2y))+(−2e^{(−x²−y²)})(0)

= 4xye^{(−x²−y²)}

f_yy(x,y) = (∂/∂y)f_y(x,y)

f_yy(x,y) = (∂/∂y)(−2ye^{(−x²−y²)})

= ((-2y)e^{(−x²−y²)}(−0−2y))+(−2e^{(−x²−y²)})(1)

= ((4y²)e^{(−x²−y²)})+(−2e^{(−x²−y²)})

= ((4y²-2)e^{(−x²−y²)})

= 2(2y²-1)e^{(−x²−y²)}

Solution :

First partial derivatives :

f_x(x,y) = −2xe^{(−x²−y²)}

f_y(x,y) = −2ye^{(−x²−y²)}

f_x(0,0) = 0

f_y(0,0) = 0

The second partial derivatives :

f_xx(x,y) = 2(2x²-1)e^{(−x²−y²)}

f_xy(x,y) = 4xye^{(−x²−y²)}

f_yy(x,y) =  2(2y²-1)e^{(−x²−y²)}