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\((\sqrt x)\) . \(\frac {1}{2\sqrt x}\)
                                = \(\frac {sec\  (tan\sqrt x) . tan\ (tan\sqrt x) . sec^2\sqrt x}{2\sqrt x}\)