Question
Answers

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) \[h\left( g\left( 3 \right) \right)=36\]

Answer Verified Verified
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.
Bookmark added to your notes.
View Notes
×