Find dy/dx of y = sin (ax + b)

The correct option is a.cos (ax + b).

To find the derivative of the function $y=\sin (ax+b)$ with respect to $x$, we apply the chain rule of differentiation.

The chain rule states that if we have a composite function $y=f\left(g\right(x\left)\right)$, its derivative is given by:
$$\frac{dy}{dx}=f^{\prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)$$

Here, the outer function is $f\left(u\right)=\sin \left(u\right)$ and the inner function is $u=g\left(x\right)=ax+b$.

First, we differentiate the outer function with respect to the inner function $u$:
$$\frac{d}{du}(\sin (u\left)\right)=\cos \left(u\right)=\cos (ax+b)$$

Next, we differentiate the inner function $u=ax+b$ with respect to $x$:
$$\frac{d}{dx}(ax+b)=a$$

Now, multiplying these two derivatives together according to the chain rule yields:
$$\frac{dy}{dx}=\cos (ax+b)\cdot a$$

Rearranging the terms, we get:
$$\frac{dy}{dx}=a\cdot \cos (ax+b)$$