$\frac{dx^5}{dx^2},$ the derivative of $x^5$ with respect to $x^2$ 
Let $f(x)=x^5.$ Find its derivative with respect to $x^2$ , i.e., find $\frac{dx^5}{dx^2}.$

We know that $\frac{dx^5}{dx}=5x^4.$
But what should I do when it is needed to take derivative with respect to $x^2 ,x^3$ or other weird things different from classical derivatives?
My approaches:
$1-)$ Say $x^2=a$ , then $x^5 =a^{5/2}$. Hence , find $$\frac{da^{5/2}}{da}=\frac{5}{2}a^{3/2}=\frac{5}{2}x^{3}$$
$2-)$ Find $$\lim_{h\to 0}\frac{f(x^2+h)-f(x^2)}{h}=\lim_{h\to 0}\frac{(x^2+h)^5-(x^2)^5}{h}=5x^8$$
My approaches conflict. Why do they give different results?
Addendum: Extra examples will be appreciated, for example, $dx^7 /dx^3,\;dx^6/dx^5.$
 A: Here is how a Physicist would do it
$$\frac{dx^5}{dx^2}\frac{dx^2}{dx}=\frac{dx^5}{dx}=5x^4$$
Since $\frac{dx^2}{dx}=2x$, we get
$$\frac{dx^5}{dx^2}=\frac{5x^3}{2}$$
A: In you 2nd attempt is wrong as stated
What is $\frac{dx^5}{dx^2}$
It is ratio of change in $ x^5$ to change in $x^2$
So $$\frac{dx^5}{dx^2}=\lim_{h\to 0}\frac{(x+h)^5-x^5}{(x+h)^2-x^2}
=\frac{5x^4h}{2xh}=\frac{5}{2}x^3$$
For general $\frac{dx^n}{dx^m}$
$$\frac{dx^m}{dx^n}=\lim_{h\to 0}\frac{(x+h)^m-x^m}{(x+h)^n-x^n}
=\frac{mx^{m-1}h}{nx^{n-1}h}=\frac{m}{n}x^{m-n}$$
And for your second approach $x^5$ is not a function of $x^2$
A: Your second attempt can't be right: it contains $f(x^2)$, which surely has no influence on the value of $\frac{df(x)}{d(x^2)}$. You have to go back to what the expression
$$\frac{df(x)}{dg(x)}$$
means. It is (roughly speaking) the amount by which $f(x)$ changes divided by the amount by which $g(x)$ changes. If $x$ changes by $h$, then $f(x)$ changes by (roughly) $hf'(x)$, and $g(x)$ changes by (roughly) $hg'(x)$. So in the limit, the expression is equal to
$$\frac{f'(x)}{g'(x)}$$
which in your case is $\frac{5x^4}{2x}=\frac52x^3$. Your extra examples are easy to evaluate using this approach.
A: In your second approach you should substitute as follows:
(note:in my solution $f(x) \neq x^5$)
$\frac{dx^5}{dx^2}=\frac{d((x^2)^{\frac{5}{2}})}{dx^2}$
So $f(x^2)=(x^2)^{\frac{5}{2}}$ which implies $f(x)=x^{\frac{5}{2}} $
\begin{align*} \lim_{h\to 0}\frac{f(x^2+h)-f(x^2)}{h}&=\lim_{h\to 0}\frac{(x^2+h)^{\frac{5}{2}}-(x^2)^{\frac{5}{2}}}{h}\\&=\lim_{h\to 0}\frac{(x^2+h)^{\frac{5}{2}}-(x^2)^{\frac{5}{2}}}{h} \cdot\frac{(x^2+h)^{\frac{5}{2}}+(x^2)^{\frac{5}{2}}}{(x^2+h)^{\frac{5}{2}}+(x^2)^{\frac{5}{2}}}\\&=\lim_{h\to 0}\frac{1}{(x^2+h)^{\frac{5}{2}}+(x^2)^{\frac{5}{2}}}\cdot\lim_{h\to 0}\frac{(x^2+h)^5-(x^2)^5}{h}\\&=\frac{1}{2x^5}\cdot\lim_{h\to 0}\frac{(x^2+h)^5-(x^2)^5}{h}\\&=\frac{1}{2x^5}\cdot5x^8\\&=\frac{5}{2}x^3
\end{align*}
