Question

Найти производную функции Z=ln(8x²+2y)

Answer 1

$\large \\ z=\ln{(8x^2+2y)}\\ \frac{\partial z}{\partial x}={1\over8x^2+2y}\cdot(16x)={8x\over4x^2+y}\\\\ \frac{\partial z}{\partial y}={1\over8x^2+2y}\cdot2={1\over4x^2+y}\\\\ \frac{\partial^2 z}{\partial x^2}={8(4x^2+y)-8x\cdot8x\over(4x^2+y)^2}={8y-32x^2\over(4x^2+y)^2 }\\\\ \frac{\partial^2 z}{\partial y^2}={0-1\over(4x^2+y)^2}=-{1\over(4x^2+y)^2}\\\\ \frac{\partial^2 z}{\partial y\partial x}={0-8x\over(4x^2+y)^2}=-{8x\over(4x^2+y)^2}\\\\ \frac{\partial^2 z}{\partial x\partial y}={0-8x\over(4x^2+y)^2}=-{8x\over(4x^2+y)^2}$