Question

$\huge\boxed{\dfrac{\partial }{\partial \:x}\left(\sqrt{x^2+y^2}\right)}$

solve it

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\bf{\dfrac{\partial }{\partial \:x}\left(\sqrt{x^2+y^2}\right)}$

♣ ᴀɴꜱᴡᴇʀ :

$\bf{Treat\:y\:as\:a\:constant}$

$\sf{Apply\:the\:chain\:rule}:\quad \dfrac{1}{2\sqrt{x^2+y^2}}\dfrac{\partial \:}{\partial \:x}\left(x^2+y^2\right)}$

$\sf{=\dfrac{1}{2\sqrt{x^2+y^2}}\dfrac{\partial \:}{\partial \:x}\left(x^2+y^2\right)}$

$\sf{\dfrac{\partial \:}{\partial \:x}\left(x^2+y^2\right)=2x}$

$\sf{=\dfrac{1}{2\sqrt{x^2+y^2}}\cdot \:2x}$

$\mathrm{Simplify\:}\dfrac{1}{2\sqrt{x^2+y^2}}\cdot \:2x:\quad \dfrac{x}{\sqrt{x^2+y^2}}$

$\huge\boxed{\bf{=\dfrac{x}{\sqrt{x^2+y^2}}}}$

Answer 2

$\mathrm{Apply\:the\:chain\:rule}:\quad \frac{1}{2\sqrt{x^2+y^2}}\frac{\partial \:}{\partial \:x}\left(x^2+y^2\right)$

$\frac{\partial \:}{\partial \:x}\left(x^2+y^2\right)=2x$

$\mathrm{Simplify\:}\frac{1}{2\sqrt{x^2+y^2}}\cdot \:2x:\quad \frac{x}{\sqrt{x^2+y^2}}$

$=\frac{x}{\sqrt{x^2+y^2}}$