Calculate each function's gradient and then evaluate it at the current point P.g(x,y)=x2−4x2y−8xy2.P(−1,2)

Step 1 The function is given as, g(x,y)=x2−4x2y−8xy2 To find the gradient of the given function at (−1 , 2). The gradient is just the vector of partial derivatives, the partial derivatives will be as, ∂g∂x=2x−8xy−8y2 and ∂g∂y=−4x2−16xy At point (−1,2), ∂g∂x=2(−1)−8(−1)(2)−8(2)2 =-2+16-32 =-18 Step 2 and ∂g∂y=−4(−12)−16(−1)(2) =-4+32 =28 Since, the gradient is represented in vector form as, ∂g∂xi+∂g∂yj, gradient =∂g∂xi+∂g∂yj =-18i+28j Hence, the gradient for the given function is −18i+28j.

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generals336

Answered question

2021-05-07

Calculate each function's gradient and then evaluate it at the current point P.
$g (x, y) = x^{2} - 4 x^{2} y - 8 x y^{2} . P (- 1, 2)$

Answer & Explanation

Cullen

Skilled2021-05-08Added 89 answers

Step 1
The function is given as,

g (x, y) = x^{2} - 4 x^{2} y - 8 x y^{2}

To find the gradient of the given function at (−1 , 2).
The gradient is just the vector of partial derivatives, the partial derivatives will be as,

\frac{\partial g}{\partial x} = 2 x - 8 x y - 8 y^{2}

and

\frac{\partial g}{\partial y} = - 4 x^{2} - 16 x y

At point (−1,2),

\frac{\partial g}{\partial x} = 2 (- 1) - 8 (- 1) (2) - 8 {(2)}^{2}

=-2+16-32
=-18
Step 2
and

\frac{\partial g}{\partial y} = - 4 (- 1^{2}) - 16 (- 1) (2)

=-4+32
=28
Since, the gradient is represented in vector form as,

\frac{\partial g}{\partial x} i + \frac{\partial g}{\partial y} j

,
gradient

= \frac{\partial g}{\partial x} i + \frac{\partial g}{\partial y} j

=-18i+28j
Hence, the gradient for the given function is −18i+28j.

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