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)