Funktion $$$\sqrt{1 - x^{2}}$$$ derivaatta

Funktio $$$\sqrt{1 - x^{2}}$$$ on kahden funktion $$$f{\left(u \right)} = \sqrt{u}$$$ ja $$$g{\left(x \right)} = 1 - x^{2}$$$ yhdistelmä $$$f{\left(g{\left(x \right)} \right)}$$$.

Sovella ketjusääntöä $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\sqrt{1 - x^{2}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right) \frac{d}{dx} \left(1 - x^{2}\right)\right)}$$

Sovella potenssisääntöä $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$, kun $$$n = \frac{1}{2}$$$:

$${\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right)\right)} \frac{d}{dx} \left(1 - x^{2}\right) = {\color{red}\left(\frac{1}{2 \sqrt{u}}\right)} \frac{d}{dx} \left(1 - x^{2}\right)$$

Palaa alkuperäiseen muuttujaan:

$$\frac{\frac{d}{dx} \left(1 - x^{2}\right)}{2 \sqrt{{\color{red}\left(u\right)}}} = \frac{\frac{d}{dx} \left(1 - x^{2}\right)}{2 \sqrt{{\color{red}\left(1 - x^{2}\right)}}}$$

Summan/erotuksen derivaatta on derivaattojen summa/erotus:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(1 - x^{2}\right)\right)}}{2 \sqrt{1 - x^{2}}} = \frac{{\color{red}\left(\frac{d}{dx} \left(1\right) - \frac{d}{dx} \left(x^{2}\right)\right)}}{2 \sqrt{1 - x^{2}}}$$

Sovella potenssisääntöä $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$, kun $$$n = 2$$$:

$$\frac{- {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(1\right)}{2 \sqrt{1 - x^{2}}} = \frac{- {\color{red}\left(2 x\right)} + \frac{d}{dx} \left(1\right)}{2 \sqrt{1 - x^{2}}}$$

Vakion derivaatta on $$$0$$$:

$$\frac{- 2 x + {\color{red}\left(\frac{d}{dx} \left(1\right)\right)}}{2 \sqrt{1 - x^{2}}} = \frac{- 2 x + {\color{red}\left(0\right)}}{2 \sqrt{1 - x^{2}}}$$

Näin ollen, $$$\frac{d}{dx} \left(\sqrt{1 - x^{2}}\right) = - \frac{x}{\sqrt{1 - x^{2}}}$$$.