Question

Differentiate the functions with respect to x.
$sec(tan(\sqrt x))$

Answer 1

Let f(x) = sec (tan($\sqrt x$), u(x) = $\sqrt x$, and v(t) = tan t and w(s) = sec s
Then, (wovou)(x) = w[v(u(x))] = w[v(√x)] = w(tan($\sqrt x$)) = sec (tan($\sqrt x$)) = f(x)
Thus, f is a composite function of three functions u, v, and w.
Put s = v(t) = tan and t = u(x) = $\sqrt x$

Then, we obtain
$\frac {dw}{ds}$ = $\frac {d}{ds}$ (sec s) = sec s.tan s = sec (tan t) . tan (tan t) [s = tan t]
= sec (tan$\sqrt x$) .tan (tan$\sqrt x$)
$\frac {ds}{dt}$ = $\frac {d}{dt}$(tan t) = sec2t = sec2$\sqrt x$
$\frac {dt}{dx}$ = $\frac {dt}{dx}$($\sqrt x$) = $\frac {dt}{dx}$($x^{\frac 12}$) = $\frac 12$ . $x^{\frac {1}{2}-1}$ = $\frac {1}{2\sqrt x}$
Therefore by chain rule, $\frac {dt}{dx}$ = $\frac {dw}{ds}$ . $\frac {ds}{dt}$ . $\frac {dt}{dx}$
= sec (tan$\sqrt x$) . tan (tan$\sqrt x$) . sec2$\sqrt x$ . $\frac {1}{2\sqrt x}$
= $\frac {1}{2\sqrt x}$ sec2$\sqrt x$ . sec (tan$\sqrt x$) . tan (tan$\sqrt x$)
= $\frac {sec\ (tan\sqrt x) . tan\ (tan\sqrt x) . sec^2\sqrt x}{2\sqrt x}$

Alternate method:

$\frac {d}{dx}$ [sec(tan($\sqrt x$))] = sec (tan$\sqrt x$) . tan (tan$\sqrt x$) . $\frac {d}{dx}$ (tan$\sqrt x$)
= sec (tan$\sqrt x$) . tan (tan$\sqrt x$) . sec2$\sqrt x$ . \frac {d}{dx}$$(\sqrt x)
= sec (tan$\sqrt x$) . tan (tan$\sqrt x$) . sec2$(\sqrt x)$ . $\frac {1}{2\sqrt x}$
= $\frac {sec\ (tan\sqrt x) . tan\ (tan\sqrt x) . sec^2\sqrt x}{2\sqrt x}$