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
ANSWERED
asked 2021-05-07
Compute the gradient of the following functions and evaluate it at the given point P.
\(\displaystyle{g{{\left({x},{y}\right)}}}={x}^{{{2}}}-{4}{x}^{{{2}}}{y}-{8}{x}{y}^{{{2}}}.{P}{\left(-{1},{2}\right)}\)

Answers (1)

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.
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