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Finding the first partial derivative and evaluating them?

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Let f(x,y) = e^[(−x^2)−1y^2]

Find the first partial derivatives fx(x,y) and fy(x,y)

Evaluate fx(0,0) and fy(0,0).

Find the second partial derivatives fxx(x,y), fxy(x,y) and fyy(x,y):
asked Nov 1, 2014 in CALCULUS by anonymous

1 Answer

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Given function f(x,y) = e(−x²−y²)

Step 1 :

First partial derivatives : fx(x,y) and fy(x,y)

fx(x,y) = (∂/∂x)f(x,y)

fx(x,y) = (∂/∂x)e(−x²−y²)

= e(−x²−y²)(∂/∂x)(−x²−y²)

= e(−x²−y²)(−2x−0)

= −2xe(−x²−y²)

fy(x,y) = (∂/∂y)f(x,y)

fy(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 : fx(0,0) and fy(0,0)

fx(x,y) = −2xe(−x²−y²)

Substitute x=0 , y=0

fx(0,0) = −2(0)e(−0²−0²)

fx(0,0) = 0

fy(x,y) = −2ye(−x²−y²)

Substitute x=0 , y=0

fy(0,0) = −2(0)e(−0²−0²)

fy(0,0) = 0

Step 3 :

The second partial derivatives : fxx(x,y),fyy(x,y) andfxy(x,y)

fxx(x,y) = (∂/∂x)fx(x,y)

fxx(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)

fxy(x,y) = (∂/∂y)fx(x,y)

fxy(x,y) = (∂/∂y)(−2xe(−x²−y²))

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

= 4xye(−x²−y²)

fyy(x,y) = (∂/∂y)fy(x,y)

fyy(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 :

fx(x,y) = −2xe(−x²−y²)

fy(x,y) = −2ye(−x²−y²)

fx(0,0) = 0

fy(0,0) = 0

The second partial derivatives :

fxx(x,y) = 2(2x²-1)e(−x²−y²)

fxy(x,y) = 4xye(−x²−y²)

fyy(x,y) =  2(2y²-1)e(−x²−y²)

answered Nov 1, 2014 by lilly Expert

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