## Calculus (3rd Edition)

$$z_x= \frac{y^2}{( x^2+y^2)^{3/2}},$$ $$z_y= \frac{ -xy}{( x^2+y^2)^{3/2}}.$$
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(x^n)'=nx^{n-1}$ Since $z= \frac{x}{\sqrt{x^2+y^2}}$, then by using the quotient rule, we have $$z_x= \frac{\sqrt{x^2+y^2} (1)-x\frac{2x}{2\sqrt{x^2+y^2}}}{ x^2+y^2}= \frac{x^2+y^2-x^2}{( x^2+y^2)^{3/2}}, \\ =\frac{y^2}{( x^2+y^2)^{3/2}},$$ $$z_y= \frac{\sqrt{x^2+y^2} (0)-x\frac{2y}{2\sqrt{x^2+y^2}}}{ x^2+y^2}= \frac{ -xy}{( x^2+y^2)^{3/2}}.$$