Question

найти частные производные функции z=√(2x+3x²y+y) в точке А(1;1)

Answer 1

$z=\sqrt{2x+3x^2y+y}$

Найдем частную производную по "х":

$\dfrac{\partial z}{\partial x} =\dfrac{1}{2\sqrt{2x+3x^2y+y}} \cdot(2x+3x^2y+y)'_x$

$\dfrac{\partial z}{\partial x} =\dfrac{1}{2\sqrt{2x+3x^2y+y}} \cdot(2+3y\cdot 2x)$

$\dfrac{\partial z}{\partial x} =\dfrac{2+6xy}{2\sqrt{2x+3x^2y+y}}$

$\dfrac{\partial z}{\partial x} (A)=\dfrac{2+6\cdot1\cdot1}{2\sqrt{2\cdot1+3\cdot1^2\cdot1+1}} =\dfrac{2+6}{2\sqrt{2+3+1}} =\dfrac{8}{2\sqrt{6}} =\dfrac{4}{\sqrt{6}}$

Найдем частную производную по "у":

$\dfrac{\partial z}{\partial y} =\dfrac{1}{2\sqrt{2x+3x^2y+y}} \cdot(2x+3x^2y+y)'_y$

$\dfrac{\partial z}{\partial y} =\dfrac{1}{2\sqrt{2x+3x^2y+y}} \cdot(3x^2+1)$

$\dfrac{\partial z}{\partial y} =\dfrac{3x^2+1}{2\sqrt{2x+3x^2y+y}}$

$\dfrac{\partial z}{\partial y} (A)=\dfrac{3\cdot1^2+1}{2\sqrt{2\cdot1+3\cdot1^2\cdot1+1}}=\dfrac{3+1}{2\sqrt{2+3+1}}=\dfrac{4}{2\sqrt{6}}=\dfrac{2}{\sqrt{6}}$