I was trying to calculate the 2nd order partial derivative $f_{xy}$ of a function defined as $$f(x,y)=\begin{cases}(x^4+y^4)tan^{-1}(\frac{y^2}{x^2}),& \text{if x$\neq$0} \\ \pi y^4/2, & \text{if x=0}\end{cases}$$
Now when trying to find the existence of the y derivative of this function at (h,0), I had a limit such as this $$f_y(h,0)=\lim_{h\rightarrow0}\frac{f(h,t)-f(h,0)}{t} \text{,where f(h,0) will be 0}$$ after substituting the function in the appropriate places, my expression was aimply reduced to $$f_y(h,0)=\lim_{t\rightarrow0}\frac{(h^4+t^4){\color{red}{tan^{-1}}\color{red}{(\frac{t^2}{h^2})}}}{\color{red}t}------(1)$$ Here, I was wondering if I could apply the L'Hospital's rule only on the red part of the equation or if I could write it as 1 as it is of the form $\lim_{x\rightarrow 0}\frac{tan^{-1}x}{x}$. Also if this from were to be applicable do I have to multiply and divide eqn (1) by $\frac{t}{h^2}$ and then proceed on with the calculation
To evaluate the limit $(1):$
$\frac{(h^4+t^4) \tan^{-1}{(\frac{t^2}{h^2})}}{t}=\frac{h^4 \tan^{-1}{(\frac{t^2}{h^2})}}{t}+t^3 \tan^{-1}{(\frac{t^2}{h^2})}$
You can aplly L'Hospital's rule to $$\frac{h^4 \tan^{-1}{(\frac{t^2}{h^2})}}{t}$$
bacause is an expression $\frac{0}{0}$ as $t \rightarrow 0$