z=ln(arctan(x^2-y^2))

Đáp án:

$dz = \dfrac{2xdx - 2ydy}{[1+ (x^2 - y^2)]\arctan(x^2 - y^2)}$

Giải thích các bước giải:

$z =\ln\left(\arctan(x^2 - y^2)\right)$

Ta có:

$+)\quad {f'}_x = \dfrac{[\arctan(x^2 - y^2)]'}{\arctan(x^2 - y^2)}$

$\to {f'}_x = \dfrac{\dfrac{(x^2 - y^2)'}{1 +(x^2 - y^2)^2}}{\arctan(x^2 - y^2)}$

$\to {f'}_x = \dfrac{2x}{[1+ (x^2 - y^2)]\arctan(x^2 - y^2)}$

$+)\quad {f'}_y= \dfrac{[\arctan(x^2 - y^2)]'}{\arctan(x^2 - y^2)}$

$\to {f'}_y = \dfrac{\dfrac{(x^2 - y^2)'}{1 +(x^2 - y^2)^2}}{\arctan(x^2 - y^2)}$

$\to {f'}_y=- \dfrac{2y}{[1+ (x^2 - y^2)]\arctan(x^2 - y^2)}$

Vậy $dz = \dfrac{2x}{[1+ (x^2 - y^2)]\arctan(x^2 - y^2)}dx - \dfrac{2y}{[1+ (x^2 - y^2)]\arctan(x^2 - y^2)}dy$

Hay $dz = \dfrac{2xdx - 2ydy}{[1+ (x^2 - y^2)]\arctan(x^2 - y^2)}$