Question

Find $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}$ at the point $(6,7)$ given $f(x,y)=\frac{x}{\sqrt{x^2+y^2}}$

Answer 1

$\begin{array}{c}Wehave\\ f\left(x,y\right)=\frac{x}{\sqrt{x^2+y^2}}\\ \frac{\partial f}{\partial x}=\frac{\sqrt{x^2+y^2}-x\frac{x}{2\sqrt{x^2+y^2}}}{\left(x^2+y^2\right)}\\ \frac{\partial f}{\partial x}=\frac{\left(x^2+y^2\right)-x^2}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}=\frac{y^2}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}\\ \therefore {\frac{\partial f}{\partial x}}_{\left(6,7\right)}=\frac{49}{{\left(85\right)}^{\frac{3}{2}}}\\ \frac{\partial f}{\partial x}=\frac{-1}{2}\frac{x\times 2y}{{\left(x^2+y^2\right)}^{\frac{3}{2}}}\\ {\frac{\partial f}{\partial x}}_{\left(6,7\right)}=\frac{-42}{{\left(85\right)}^{\frac{3}{2}}}\end{array}$