Differentiate the functions with respect to x.

Let $f(x)=\sin (a x+b), u(x)=a x+b$, and $v(t)=\sin t$

Then, $($ vou $)(x)=v(u(x))=v(a x+b)=\sin (a x+b)=f(x)$

Thus, f is a composite function of two functions, u and v.

Put t = u (x) = ax + b

Therefore,

$\frac{d v}{d t}=\frac{d}{d t}(\sin t)=\cos t=\cos (a x+b)$

$\frac{d t}{d x}=\frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)=a+0=a$

Hence, by chain rule, we obtain

$\frac{d f}{d x}=\frac{d v}{d t} \cdot \frac{d t}{d x}=\cos (a x+b) \cdot a=a \cos (a x+b)$

Alternate method

$\frac{d}{d x}[\sin (a x+b)]=\cos (a x+b) \cdot \frac{d}{d x}(a x+b)$

$=\cos (a x+b) \cdot\left[\frac{d}{d x}(a x)+\frac{d}{d x}(b)\right]$

$=\cos (a x+b) \cdot(a+0)$

$=a \cos (a x+b)$