Question

# Let $f:R\to R,\text{ }g:R\to R\text{ and }h:R\to R$be differentiable functions such that $f\left( x \right)={{x}^{3}}+3x+2,\text{ }g\left( f\left( x \right) \right)=x\text{ and }h\left( g\left( g\left( x \right) \right) \right)=x\text{ for all }x\in R.$Then.(a) ${{g}^{'}}\left( 2 \right)=\dfrac{1}{15}$(b) ${{h}^{'}}\left( 1 \right)=666$(c) $h\left( 0 \right)=16$(d) $h\left( g\left( 3 \right) \right)=36$

Hint: Differentiate $g\left( f\left( x \right) \right)=x\text{ and }h\left( g\left( g\left( x \right) \right) \right)=x$by replacing $x\text{ by }f\left( x \right)\text{in }h\left( g\left( g\left( x \right) \right) \right)=x$twice.

We are given that $f\left( x \right)={{x}^{3}}+3x+2,\text{ }g\left( f\left( x \right) \right)=x\text{ and }h\left( g\left( g\left( x \right) \right) \right)=x\text{ for }x\in R.$
Now, we have to find the values of ${{g}^{'}}\left( 2 \right),{{h}^{'}}\left( 1 \right),h\left( 0 \right)\text{ and }h\left( g\left( 3 \right) \right)$
We are given that $g\left( f\left( x \right) \right)=x$
Now, we will differentiate both sides with respect to x.
Also we know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
Also, by chain rule, if $y=f\left( u \right)\text{ and }u=g\left( x \right)$, then
$\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$
Therefore, we get ${{g}^{'}}\left( f\left( x \right) \right).{{f}^{'}}\left( x \right)=1$
${{g}^{'}}\left( f\left( x \right) \right)=\dfrac{1}{{{f}^{'}}\left( x \right)}....\left( i \right)$
Also, $f\left( x \right)={{x}^{3}}+3x+2$
By differentiating both sides with respect to x,
We get, ${{f}^{'}}\left( x \right)=3{{x}^{2}}+3....\left( ii \right)$
Now, to get $g'\left( 2 \right)$, we put $f\left( x \right)=2$in equation (i)
We get, ${{g}^{'}}\left( 2 \right)=\dfrac{1}{{{f}^{'}}\left( x \right)}....\left( iii \right)$
Now, for $f\left( x \right)=2$
That is ${{x}^{3}}+3x+2=2$
We get, ${{x}^{3}}+3x=0$
$x\left( {{x}^{2}}+3 \right)=0$
As we know that ${{x}^{2}}\ne -3$, therefore we get x = 0
By putting x = 0 in equation (iii), we get
${{g}^{'}}\left( 2 \right)=\dfrac{1}{{{f}^{'}}\left( 0 \right)}$
To get ${{f}^{'}}\left( 0 \right)$, we put x = 0 in equation (ii)
We get ${{f}^{'}}\left( 0 \right)=3$
Therefore we get, ${{g}^{'}}\left( 2 \right)=\dfrac{1}{3}$
Now we are given that $h\left( g\left( g\left( x \right) \right) \right)=x$
Here we will replace x by $f\left( x \right)$,
So we get $h\left( g\left( g\left( f\left( x \right) \right) \right) \right)=f\left( x \right)$
We know that $g\left( f\left( x \right) \right)=x$
$h\left( g\left( x \right) \right)=f\left( x \right)....\left( iv \right)$
Now, again we will replace x by $f\left( x \right)$.
So we get, $h\left( g\left( f\left( x \right) \right) \right)=f\left( f\left( x \right) \right)$
We know that $g\left( f\left( x \right) \right)=x$
So, we get $h\left( x \right)=f\left( f\left( x \right) \right)....\left( v \right)$
Now by differentiating both sides,
And by chain rule, if $y=f\left( u \right)\text{ and }u=g\left( x \right)$then
$\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$
So, we get ${{h}^{'}}\left( x \right)={{f}^{'}}\left( f\left( x \right) \right).{{f}^{'}}\left( x \right)....\left( vi \right)$
Now to get $h\left( g\left( 3 \right) \right)$, we will first put x = 3 in equation (iv)
So we get $h\left( g\left( 3 \right) \right)=f\left( 3 \right)$
Therefore, $h\left( g\left( 3 \right) \right)={{\left( 3 \right)}^{3}}+3\left( 3 \right)+2=38$
Now to get $h\left( 0 \right)$, we will put x = 0 in equation (v)
We get $h\left( 0 \right)=f\left( f\left( 0 \right) \right)$
$\Rightarrow h\left( 0 \right)=f\left( {{\left( 0 \right)}^{3}}+3\left( 0 \right)+2 \right)$
$\Rightarrow h\left( 0 \right)=f\left( 2 \right)$
$\Rightarrow h\left( 0 \right)={{\left( 2 \right)}^{3}}+3\left( 2 \right)+2$
Therefore, we get $h\left( 0 \right)=8+6+2=16$
Now, to get ${{h}^{'}}\left( 1 \right)$, we will put x = 1 in equation (vi)
${{h}^{'}}\left( 1 \right)={{f}^{'}}\left( f\left( 1 \right) \right).{{f}^{'}}\left( 1 \right)$
$={{f}^{'}}\left( {{\left( 1 \right)}^{3}}+3\left( 1 \right)+2 \right).{{f}^{'}}\left( 1 \right)$
$={{f}^{'}}\left( 6 \right).{{f}^{'}}\left( 1 \right)$
To get ${{f}^{'}}\left( 6 \right)$and ${{f}^{'}}\left( 1 \right)$, we will put x = 6 and x = 1 in equation (ii)
We get, ${{h}^{'}}\left( 1 \right)=\left[ 3{{\left( 6 \right)}^{2}}+3 \right].\left[ 3{{\left( 1 \right)}^{2}}+3 \right]$
$=\left( 3\times 36+3 \right).\left( 6 \right)$
$=\left( 111 \right).6$
$=666$
Therefore we get,
\begin{align} & \Rightarrow {{g}^{'}}\left( 2 \right)=\dfrac{1}{3} \\ & \Rightarrow {{h}^{'}}\left( 1 \right)=666 \\ & \Rightarrow h\left( 0 \right)=16 \\ & \Rightarrow h\left( g\left( 3 \right) \right)=38 \\ \end{align}
Hence option (b) and (c) are correct.

Note: Students often write differentiation of $g\left( f\left( x \right) \right)={{g}^{'}}f\left( x \right)$and miss the differentiation of $f\left( x \right)$which is ${{f}^{'}}\left( x \right)$. So they must keep in mind the chain rule and correct differentiation of $g\left( f\left( x \right) \right)$is ${{g}^{'}}f\left( x \right).{{f}^{'}}\left( x \right)$and similar rules must be followed for all composite functions.