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)