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

Last updated date: 26th Mar 2023

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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.

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.

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