Question

Differentiate the functions with respect to x.
$sin\ (ax+b)$

Answer 1

Let f(x) = sin (ax+b), u(x) = ax+b, and v(t) = sint
Then, (vou)(x) = v(u(x)) = v(ax+b) = sin(ax+b) = f(x)
Thus, f is a composite of two functions.
Put t = u(x) = ax+b

Then, we obtain
$\frac {dv}{dt}$=$\frac {d}{dt}$(sin t) = cost = cos (ax+b)
$\frac {dt}{dx}$=$\frac {d}{dx}$(ax+b) = $\frac {d}{dx}$(ax) + $\frac {d}{dx}$(b) = a+0 = a
Therefore by chain rule, $\frac {df}{dx}$=$\frac {dv}{dt}$.$\frac {dt}{dx}$ = cos (ax+b) . a = acos (ax+b)

Alternate method:

$\frac {d}{dx}$[sin (ax+b)] = cos (ax+b) . $\frac {d}{dx}$(ax+b)
=cos (ax+b) . [$\frac {d}{dx}$(ax) + $\frac {d}{dx}$(b)]
=cos (ax+b) . [a+0]
=a cos (ax+b)