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

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

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

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

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

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

Palaa alkuperäiseen muuttujaan:

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

Summan/erotuksen derivaatta on derivaattojen summa/erotus:

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

Vakion derivaatta on $$$0$$$:

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

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

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

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