Question

If \[f\left( x \right) = \dfrac{x}{{\left( {1 + x} \right)}}\] , and \[g\left( x \right) = f\left[ {f\left( x \right)} \right]\]. Then what is the value of \[g'\left( x \right)\]?
A. \[\dfrac{1}{{{{\left( {2x + 3} \right)}^2}}}\]
B. \[\dfrac{1}{{{{\left( {x + 1} \right)}^2}}}\]
C. \[\dfrac{1}{{{x^2}}}\]
D. \[\dfrac{1}{{{{\left( {2x + 1} \right)}^2}}}\]

Answer 1

Hint In the given question, two functions are given. By substituting the value of first function in the second composite function, we will find the value of \[g\left( x \right)\]. Then by differentiating the function \[g\left( x \right)\] with respect to \[x\], we will find the value of \[g'\left( x \right)\].

Formula used
Composite function: A function is called a composite function when one function is substituted into another function.

Quotient rule of differentiation: \[\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}f\left( x \right) - f\left( x \right)\dfrac{d}{{dx}}g\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\]


Complete step by step solution:
The given functions are \[f\left( x \right) = \dfrac{x}{{\left( {1 + x} \right)}}\] and \[g\left( x \right) = f\left[ {f\left( x \right)} \right]\].
Let’s simplify the given composite function.
\[g\left( x \right) = f\left[ {f\left( x \right)} \right]\]
Substitute \[f\left( x \right) = \dfrac{x}{{\left( {1 + x} \right)}}\] in above function.
\[g\left( x \right) = f\left[ {\dfrac{x}{{\left( {1 + x} \right)}}} \right]\]
Apply the rule of the function \[f\left( x \right)\]
\[g\left( x \right) = \dfrac{{\left( {\dfrac{x}{{\left( {1 + x} \right)}}} \right)}}{{\left( {1 + \dfrac{x}{{\left( {1 + x} \right)}}} \right)}}\]
Now simplify the above function.
\[g\left( x \right) = \dfrac{{\left( {\dfrac{x}{{\left( {1 + x} \right)}}} \right)}}{{\left( {\dfrac{{1 + x + x}}{{\left( {1 + x} \right)}}} \right)}}\]
\[ \Rightarrow \]\[g\left( x \right) = \dfrac{x}{{1 + 2x}}\]
Differentiate the above function with respect to \[x\].
Apply quotient rule of differentiation \[\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}f\left( x \right) - f\left( x \right)\dfrac{d}{{dx}}g\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\] on the right-hand side.
\[g'\left( x \right) = \dfrac{{\left( {1 + 2x} \right)\dfrac{d}{{dx}}\left( x \right) - x\dfrac{d}{{dx}}\left( {1 + 2x} \right)}}{{{{\left( {1 + 2x} \right)}^2}}}\]
\[ \Rightarrow \]\[g'\left( x \right) = \dfrac{{\left( {1 + 2x} \right)\left( 1 \right) - x\left( 2 \right)}}{{{{\left( {1 + 2x} \right)}^2}}}\]
Now simplify the above function.
\[g'\left( x \right) = \dfrac{{1 + 2x - 2x}}{{{{\left( {1 + 2x} \right)}^2}}}\]
\[ \Rightarrow \]\[g'\left( x \right) = \dfrac{1}{{{{\left( {1 + 2x} \right)}^2}}}\]
Hence the correct option is D.

Note: Students often get confused with the quotient rule of differentiation \[\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}f\left( x \right) - f\left( x \right)\dfrac{d}{{dx}}g\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\] and \[\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) - g\left( x \right)\dfrac{d}{{dx}}f\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\]. The correct quotient rule of differentiation is \[\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)\dfrac{d}{{dx}}f\left( x \right) - f\left( x \right)\dfrac{d}{{dx}}g\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\].