Let f: R → R, g: R → R and h: R → R be differentiable functions...

If ${f}^{′}\left(x\right) = 3{x}^{2} + 3$
$⇒ g\left(f\left(x\right)\right) = x$
$⇒ {g}^{′}\left(f\left(x\right)\right) = \dfrac{1}{{f}^{′}\left(x\right)}$
Also f\left(0\right) = 2$($Put x = 0$)<br />So,${g}^{′}\left(2\right) = \dfrac{1}{{f}^{′}\left(0\right)}$<br />$⇒ g′\left(2\right) = \dfrac{1}{3}$<br />If$h ( g\left(g\left(x\right)\right) = x$<br />${h}^{′}\left(g\left(g\left(x\right)\right)\right) = \dfrac{1}{{g}^{′}\left(g\left(x\right)\right).{g}^{′}\left(x\right)}$<br />If,$g\left(g\left(x\right)\right) = 1 g ( f\left(x\right) ) = x ⇒ {g}^{- 1}\left(x\right) = f ( x )$<br />$⇒ g\left(x\right) = {g}^{- 1}\left(1\right) ⇒ g\left(x\right) = f\left(1\right) = 6$<br />To find,${g}^{'}\left(6\right)$use$x = 1$in<br />${g}^{'}\left(f\left(x\right)\right) = \dfrac{1}{{f}^{'}\left(1\right)}$as$f \left(1\right) = 6$</p><p>To find,${g}^{'}\left(236\right)$, use$x = 6$as<br />$f\left(6\right) = 236$<br />$∴ {h}^{'}\left(1\right) = \dfrac{1}{{g}^{'}\left(6\right).{g}^{'} ( 236 )} = \dfrac{1}{\dfrac{1}{6}.\dfrac{1}{111}} ⇒ {h}^{'}\left(1\right) = 666$<br />If$g\left(g\left(x\right)\right) = 0$<br />$∴ g\left(x\right) = {g}^{- 1}\left(0\right) ⇒ g\left(x\right) = f\left(0\right)$</p><p>$⇒ g\left(x\right) = 2 ⇒ x = {g}^{- 1}\left(2\right)$</p><p>$⇒ x = f\left(2\right) ⇒ x = 16$<br />$∴ h\left(0\right) = 16$<br />$g\left(x\right) = 3 ⇒ x = {g}^{- 1}\left(3\right)$</p><p>$⇒ x = f\left(3\right) ⇒ x = 38$<br />So$h\left(g\left(3\right)\right) = 38$