Derivative of $\sqrt{a^2-x^2}$

Question

Can anyone provide an explanation for the following:

$$ \frac{d}{dx}(\sqrt{a^2-x^2}) = \frac{-x}{\sqrt{a^2-x^2}} $$

I can only seem to get $\frac{a-x}{\sqrt{a^2-x^2}}$. I don't understand how the derivative of $a^2-x^2 = -2x$. Why does the $a^2$ term differentiate into nothing? I understand this exchange is for questions of a much higher caliber, but any help would be greatly appreciated. Thanks.

The derivative of a constant is $0$. – Kavi Rama Murthy Aug 02 '21 at 09:10 — Kavi Rama Murthy, Aug 02 '21 at 09:10
@KaviRamaMurthy Ohhhhhh, thank you very much. – Freddy Mcloughlan Aug 02 '21 at 09:11 — Freddy Mcloughlan, Aug 02 '21 at 09:11

score 0 · Accepted Answer · answered Aug 02 '21 at 14:53

We can apply the derivative definition of a function $y=f(x)$:

it is the limit of the incremental ratio $\Delta y$ when $h=\Delta x$ tends to zero.

Here $f(x)=\sqrt{a^{2}-x^{2}}$.

$y’=\lim\limits_{h \to 0}\frac{\Delta y}{h}$;

$\Delta y=f(x+h)-f(x)$;

$\Delta y=\sqrt{a^{2}-(x+h)^{2}}-\sqrt{a^{2}-x^{2}}$.

Multiplying and dividing by $\sqrt{a^{2}-(x+h)^{2}}+\sqrt{a^{2}-x^{2}}$, we get:

$\frac{\Delta y }{h} =\frac{a^{2}-(x+h)^{2}-(a^{2}-x^{2})} {h(\sqrt{a^{2}-(x+h)^{2}}+\sqrt{a^{2}-x^{2}})}$,

$=\frac{-2hx-h^{2}} { h(\sqrt{a^{2}-(x+h)^{2}}+\sqrt{a^{2}-x^{2}})}$,

going to the limit:

$\lim\limits_{h \to 0}\frac{\Delta y}{h}=\frac{-x}{\sqrt{a^{2}-x^{2}}}$.

Derivative of $\sqrt{a^2-x^2}$

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