<s>[INST] $\frac{dx^5}{dx^2},$ 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, $dx^7 /dx^3,\;dx^6/dx^5.$
 [/INST] 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*}
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