Question

Find the gradient of w = x^2y^3z at (1, 2,−1).

Answer 1

have you taken partial derivatives of w ?

Answer 2

\[\frac{\partial w}{\partial x} = w_x = \quad...\]\[\frac{\partial w}{\partial y} = w_y = \quad...\]\[\frac{\partial w}{\partial z} = w_z = \quad...\]

Answer 3

The gradient of a (scalar) function \(w\), is the vector that's components are the respective partial derivatives
\[\operatorname\nabla w=w_x\,\hat{\mathrm{\textbf x}}+w_y\,\hat{\mathrm{\textbf y}}+w_z\,\hat{\mathrm{\textbf z}}=(w_x,w_y,w_z)\]

Answer 4

To take the partial derivative of \(w\) with respect to \(x\)

\[\frac{\partial w}{\partial x} =\frac{\partial }{\partial x}\left(x^2y^3z\right) \] treat \(y\), and \(z\), as constants

Answer 5

i.e.\[\frac{\partial w}{\partial x} =y^3z\frac{\partial }{\partial x}\left(x^2\right)\]
...

Answer 6

Are you there @tote ?