Find the partial derivative of $\arctan (x/\sqrt{x^2+y^2})$ using the definition Let $f(x,y)=\arctan \frac{x}{\sqrt{x^2+y^2}}$. How to evaluate $$\lim_{h\to 0}\frac{f(4h,1)-f(h,1)}{h}?$$
 A: Hints: Since your limit is $0/0$ form, so use $\arctan x\sim x$.
$f(x,y)=\arctan \frac{x}{\sqrt{x^2+y^2}}$. So $f(4h,1)=\arctan \frac{4h}{\sqrt{16h^2+1}}\sim \frac{4h}{\sqrt{16h^2+1}}$ and $f(h,1)=\arctan \frac{h}{\sqrt{h^2+1}}\sim \frac{h}{\sqrt{h^2+1}}$.
A: I see no partial derivative, here. However, writing the limit in the form
$$
\lim_{h\to0}\left(
  4\cdot\frac{f(4h,1)-f(0,1)}{4h}-\frac{f(h,1)-f(0,1)}{h}
\right)
$$
we see the limit is
$$
4\frac{\partial f}{\partial x}\bigg|_{(0,1)}
-
\frac{\partial f}{\partial x}\bigg|_{(0,1)}=
3\frac{\partial f}{\partial x}\bigg|_{(0,1)}
$$
Since
\begin{align}
\frac{\partial f}{\partial x}
&=
\frac{1}{1+\dfrac{x^2}{x^2+y^2}}
\frac{\sqrt{x^2+y^2}-\dfrac{x^2}{\sqrt{x^2+y^2}}}{x^2+y^2}\\
&=
\frac{x^2+y^2}{2x^2+y^2}\frac{y^2}{(x^2+y^2)\sqrt{x^2+y^2}}
\end{align}
your limit is $3$.
Of course it's easier to consider that
$$
\frac{t}{\sqrt{t^2+1}}=t+o(t),\qquad \arctan t=t+o(t)
$$
so we can write
$$
\lim_{h\to0}\frac{f(4h,1)-f(h,1)}{h}=
\lim_{h\to0}\frac{4h+o(4h)-(h+o(h))}{h}=3
$$
