Derivative of $$$\cos{\left(b - x \right)}$$$ with respect to $$$x$$$
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Find $$$\frac{d}{dx} \left(\cos{\left(b - x \right)}\right)$$$.
Solution
The function $$$\cos{\left(b - x \right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ and $$$g{\left(x \right)} = b - x$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\cos{\left(b - x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{dx} \left(b - x\right)\right)}$$The derivative of the cosine is $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:
$${\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{dx} \left(b - x\right) = {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{dx} \left(b - x\right)$$Return to the old variable:
$$- \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(b - x\right) = - \sin{\left({\color{red}\left(b - x\right)} \right)} \frac{d}{dx} \left(b - x\right)$$The derivative of a sum/difference is the sum/difference of derivatives:
$$- \sin{\left(b - x \right)} {\color{red}\left(\frac{d}{dx} \left(b - x\right)\right)} = - \sin{\left(b - x \right)} {\color{red}\left(\frac{db}{dx} - \frac{d}{dx} \left(x\right)\right)}$$The derivative of a constant is $$$0$$$:
$$- \left({\color{red}\left(\frac{db}{dx}\right)} - \frac{d}{dx} \left(x\right)\right) \sin{\left(b - x \right)} = - \left({\color{red}\left(0\right)} - \frac{d}{dx} \left(x\right)\right) \sin{\left(b - x \right)}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$\sin{\left(b - x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = \sin{\left(b - x \right)} {\color{red}\left(1\right)}$$Thus, $$$\frac{d}{dx} \left(\cos{\left(b - x \right)}\right) = \sin{\left(b - x \right)}$$$.
Answer
$$$\frac{d}{dx} \left(\cos{\left(b - x \right)}\right) = \sin{\left(b - x \right)}$$$A