Question

Полный дифференциал Z=arctg √(x^y)

Answer 1

$z=arctg \sqrt{x^y}=arctg \sqrt{e^{y lnx}}$
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$d z=(arctg \srqt{e^{y lnx}})'_x dx+(arctg \srqt{e^{y lnx}})'_y dy=$
$=\frac{1}{1+(\sqrt{x^y})^2}*\frac{1}{2\sqrt{x^y}}*e^{y lnx}*y*\frac{1}{x} dx+$
$\frac{1}{1+(\sqrt{x^y})^2}*\frac{1}{2\sqrt{x^y}}*e^{y lnx}*ln x*1 dy=$
$\frac{x^y * y}{2x(1+x^y) *\sqrt{x^y}} dx+ \frac{x^y * lnx}{2(1+x^y) *\sqrt{x^y}} dy$
$=\frac{\sqrt{x^y}*y}{2x(1+x^y)} dx+\frac{\sqrt{x^y}*ln x}{2(1+x^y)} dy$