Question

Compute the gradient of the following functions and evaluate it at the given point P. g(x,y)=x^{2}-4x^{2}y-8xy^{2}. P(-1,2)

Derivatives
Compute the gradient of the following functions and evaluate it at the given point P.
$$\displaystyle{g{{\left({x},{y}\right)}}}={x}^{{{2}}}-{4}{x}^{{{2}}}{y}-{8}{x}{y}^{{{2}}}.{P}{\left(-{1},{2}\right)}$$

2021-05-08
Step 1
The function is given as, $$\displaystyle{g{{\left({x},{y}\right)}}}={x}^{{{2}}}-{4}{x}^{{{2}}}{y}-{8}{x}{y}^{{{2}}}$$
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,
$$\displaystyle{\frac{{\partial{g}}}{{\partial{x}}}}={2}{x}-{8}{x}{y}-{8}{y}^{{{2}}}$$ and
$$\displaystyle{\frac{{\partial{g}}}{{\partial{y}}}}=-{4}{x}^{{{2}}}-{16}{x}{y}$$
At point (−1,2),
$$\displaystyle{\frac{{\partial{g}}}{{\partial{x}}}}={2}{\left(-{1}\right)}-{8}{\left(-{1}\right)}{\left({2}\right)}-{8}{\left({2}\right)}^{{{2}}}$$
=-2+16-32
=-18
Step 2
and
$$\displaystyle{\frac{{\partial{g}}}{{\partial{y}}}}=-{4}{\left(-{1}^{{{2}}}\right)}-{16}{\left(-{1}\right)}{\left({2}\right)}$$
=-4+32
=28
Since, the gradient is represented in vector form as, $$\displaystyle{\frac{{\partial{g}}}{{\partial{x}}}}{i}+{\frac{{\partial{g}}}{{\partial{y}}}}{j}$$,
gradient $$\displaystyle={\frac{{\partial{g}}}{{\partial{x}}}}{i}+{\frac{{\partial{g}}}{{\partial{y}}}}{j}$$
=-18i+28j
Hence, the gradient for the given function is −18i+28j.