Diferential Total

[Chipset3600]

Question, is it correct??
Calculate the total differential of this function: \(\displaystyle u= arcsen\sqrt{\frac{xz}{y}}\)

1- respect of x: \(\displaystyle \frac{\partial u}{\partial x}=\frac{zy}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}\)

2- respect of y: \(\displaystyle \frac{\partial u}{\partial y}=\frac{xz}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}\)

3- respect of z: \(\displaystyle \frac{\partial u}{\partial z}=\frac{xy}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}\)

int the formula: \(\displaystyle :\partial u=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}


final: \partial u=\frac{zy}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}+\frac{xz}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}+\frac{xy}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}\)

 

[Deleted member 4993]

Question, is it correct??
Calculate the total differential of this function: \(\displaystyle u= arcsen\sqrt{\frac{xz}{y}}\)

1- respect of x: \(\displaystyle \frac{\partial u}{\partial x}=\frac{zy}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}\)

2- respect of y: \(\displaystyle \frac{\partial u}{\partial y}=\frac{xz}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}\)

3- respect of z: \(\displaystyle \frac{\partial u}{\partial z}=\frac{xy}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}\)

int the formula: \(\displaystyle :\ du=\frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy + \frac{\partial u}{\partial z} dz\)


final:\(\displaystyle \ du=\frac{zy \ dx}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}+\frac{xz \ dy}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}+\frac{xy \ dz}{{2y^{2}}\sqrt{\frac{xz}{y}\sqrt{1-\frac{xz}{y}}}}\)

Looks good - except as noted above

 

[Chipset3600]

Looks good - except as noted above

This account was worth 5 points
My teacher just gave me 2, dont know why