(log(1+(1.5-x+x*y)^2+(2.25-x+x*y^2)^2+(2.625-x+x*y^3)^2))/10求梯度
时间: 2023-10-21 19:06:14 浏览: 39
我们需要对每个变量求偏导数,然后组成梯度向量。首先,我们可以将这个函数写成更简洁的形式:
$f(x,y) = \frac{1}{10} \log(1 + (1.5 - x + xy)^2 + (2.25 - x + xy^2)^2 + (2.625 - x + xy^3)^2)$
现在我们对 $x$ 求偏导数:
$\frac{\partial f}{\partial x} = \frac{1}{10(1 + (1.5 - x + xy)^2 + (2.25 - x + xy^2)^2 + (2.625 - x + xy^3)^2)} \cdot (-2(1.5 - x + xy) - 2(2.25 - x + xy^2) - 2(2.625 - x + xy^3))$
$= \frac{-2(1.5 - x + xy) - 2(2.25 - x + xy^2) - 2(2.625 - x + xy^3)}{10(1 + (1.5 - x + xy)^2 + (2.25 - x + xy^2)^2 + (2.625 - x + xy^3)^2)}$
现在我们对 $y$ 求偏导数:
$\frac{\partial f}{\partial y} = \frac{1}{10(1 + (1.5 - x + xy)^2 + (2.25 - x + xy^2)^2 + (2.625 - x + xy^3)^2)} \cdot (2x(1.5 - x + xy) + 2xy(2.25 - x + xy^2) + 2xy^2(2.625 - x + xy^3))$
$= \frac{2x(1.5 - x + xy) + 2xy(2.25 - x + xy^2) + 2xy^2(2.625 - x + xy^3)}{10(1 + (1.5 - x + xy)^2 + (2.25 - x + xy^2)^2 + (2.625 - x + xy^3)^2)}$
因此,梯度向量为:
$\nabla f(x,y) = \begin{pmatrix} \frac{-2(1.5 - x + xy) - 2(2.25 - x + xy^2) - 2(2.625 - x + xy^3)}{10(1 + (1.5 - x + xy)^2 + (2.25 - x + xy^2)^2 + (2.625 - x + xy^3)^2)} \\ \frac{2x(1.5 - x + xy) + 2xy(2.25 - x + xy^2) + 2xy^2(2.625 - x + xy^3)}{10(1 + (1.5 - x + xy)^2 + (2.25 - x + xy^2)^2 + (2.625 - x + xy^3)^2)} \end{pmatrix}$