Easy
Differentiate the function with respect to $x$: $\sin (ax+b)$
Let $f(x) = \sin (ax+b)$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{d}{dx}[\sin (ax+b)] = \cos (ax+b) \cdot \frac{d}{dx}(ax+b)$
$= \cos (ax+b) \cdot [\frac{d}{dx}(ax) + \frac{d}{dx}(b)]$
$= \cos (ax+b) \cdot (a + 0)$
$= a \cos (ax+b)$
Using the chain rule,we differentiate with respect to $x$:
$\frac{d}{dx}[\sin (ax+b)] = \cos (ax+b) \cdot \frac{d}{dx}(ax+b)$
$= \cos (ax+b) \cdot [\frac{d}{dx}(ax) + \frac{d}{dx}(b)]$
$= \cos (ax+b) \cdot (a + 0)$
$= a \cos (ax+b)$
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