Funktion $$$\cos{\left(b - x \right)}$$$ derivaatta muuttujan $$$x$$$ suhteen

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

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

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

Kosinin derivaatta on $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:

$${\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{dx} \left(b - x\right) = {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{dx} \left(b - x\right)$$

Palaa alkuperäiseen muuttujaan:

$$- \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(b - x\right) = - \sin{\left({\color{red}\left(b - x\right)} \right)} \frac{d}{dx} \left(b - x\right)$$

Summan/erotuksen derivaatta on derivaattojen summa/erotus:

$$- \sin{\left(b - x \right)} {\color{red}\left(\frac{d}{dx} \left(b - x\right)\right)} = - \sin{\left(b - x \right)} {\color{red}\left(\frac{db}{dx} - \frac{d}{dx} \left(x\right)\right)}$$

Vakion derivaatta on $$$0$$$:

$$- \left({\color{red}\left(\frac{db}{dx}\right)} - \frac{d}{dx} \left(x\right)\right) \sin{\left(b - x \right)} = - \left({\color{red}\left(0\right)} - \frac{d}{dx} \left(x\right)\right) \sin{\left(b - x \right)}$$

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

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

Näin ollen, $$$\frac{d}{dx} \left(\cos{\left(b - x \right)}\right) = \sin{\left(b - x \right)}$$$.