Question

Differentiate the functions with respect to x.
\(sin(x^2+5)\)

Answer 1

Let f(x) = sin (x²+5), u(x) = x²+5, and v(t) = sin t
Then, (vou)x = v(u(x)) = v(x²+5) = tan(x²+5) = f(x)
Thus, f is a composite of two functions.

Put t = u(x) = x²+5
Then we obtain 
\(\frac {dv}{dt}\) = \(\frac {d}{dt}\)(sin t) = cos t = cos (x²+5)
\(\frac {dt}{dx}\) = \(\frac {d}{dt}\)(x²+5) = \(\frac {d}{dt}\)(x²) + \(\frac {d}{dt}\)(5) = 2x+0 = 2x
Therefore by chain rule, \(\frac {df}{dx}\) = \(\frac {dv}{dt}\) . \(\frac {dt}{dx}\) = cos (x²+5) . 2x = 2x cos(x²+5)

Alternate method:
\(\frac {d}{dx}\)[sin (x²+5)] = cos (x²+5) . \(\frac {d}{dx}\)(x²+5)
=cos (x²+5) . [\(\frac {d}{dx}\)(x²) + \(\frac {d}{dx}\)(5)]
=cos (x²+5) . [2x + 0]
=2x cos (x²+5)