Question

Найти частную производную первого порядка.

Answer 1

$z'_y=\frac{1}{2x^{2} -y}$

$z'_x= \frac{6x^{2} -y}{2x^{3} -xy}$

Объяснение:

$(\ ln(f(t) )\ )' _t = \frac{1}{f(t)} *f'(t)\\$

$z'_x =(\ ln(x^{3} -\frac{xy}{2} )\ )' _x = \frac{1}{x^{3} -\frac{xy}{2}} *(x^{3} -\frac{xy}{2})'_x= \frac{1}{x^{3} -\frac{xy}{2}} *(3x^{2} -\frac{y}{2})\\ = \frac{3x^{2} -\frac{y}{2}}{x^{3} -\frac{xy}{2}} = \frac{6x^{2} -y}{2x^{3} -xy}$

$z' _y =(\ ln(x^{3} -\frac{xy}{2} )\ )' _y = \frac{1}{x^{3} -\frac{xy}{2}} *(x^{3} -\frac{xy}{2})'_y= \frac{1}{x^{3} -\frac{xy}{2}} *(0 -\frac{x}{2})\\ = -\frac{x}{2x^{3} -xy} = -\frac{1}{2x^{2} -y}$