Question

If we have a function as \[y = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)\], then find \[\dfrac{{dy}}{{dx}}\].

Answer 1

Hint: We need to find the derivative of \[{\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)\]. We see that this function is the composition of two functions i.e. \[{\tan ^{ - 1}}\left( x \right)\] and \[\dfrac{{2x}}{{1 - {x^2}}}\]. And since we have to find the derivative of this composite function, we will use chain rule, which states that
\[\dfrac{d}{{dx}}\left( {fog(x)} \right) = f'(g(x)) \times g'(x)\], where \[g'(x) = \dfrac{{dg}}{{dx}}\].
Here, we again see that \[\dfrac{{2x}}{{1 - {x^2}}}\] is a fraction. To find the derivative of this function, we need to use the quotient rule. Quotient rule states that:
\[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}}\], where \[u\] and \[v\] are functions of \[x\] and \[v' = \dfrac{{dv}}{{dx}}\]

Complete step-by-step solution:
Finding the derivative of \[{\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)\]. Now, writing this function as a composition of two functions.
Let \[f(x) = {\tan ^{ - 1}}x - - - - - - (1)\]
And \[g(x) = \dfrac{{2x}}{{1 - {x^2}}} - - - - - - (2)\]
Using (1) and (2), we get
\[fog(x) = f\left( {g\left( x \right)} \right)\]
\[ \Rightarrow fog(x) = {\tan ^{ - 1}}\left( {g\left( x \right)} \right)\]
Substituting \[g(x) = \dfrac{{2x}}{{1 - {x^2}}}\] in above expression, we have
\[ \Rightarrow fog(x) = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) - - - - - - (3)\]
Now, to find the derivative of this function, we have
\[\dfrac{d}{{dx}}\left( {fog(x)} \right) = f'(g(x)) \times g'(x)\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {fog(x)} \right) = \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)} \right) = f'\left( {g\left( x \right)} \right) \times g'(x) - - - - - - (4)\]
Using (1), we will find \[f'\left( {g\left( x \right)} \right)\] first
For that we first have to find \[f'\left( x \right) = \dfrac{{df}}{{dx}}\]
From (1), we have
\[\dfrac{{df}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( x \right)} \right)\]
As we know, \[\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( x \right)} \right) = \dfrac{1}{{1 + {x^2}}}\]. Using in the above formula, we have
\[ \Rightarrow \dfrac{{df}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( x \right)} \right) = \dfrac{1}{{1 + {x^2}}}\]
\[ \Rightarrow \dfrac{{df}}{{dx}} = f'\left( x \right) = \dfrac{1}{{1 + {x^2}}}\]
Now, replacing \[x\] by \[g(x)\], we have
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{1}{{1 + {{\left( {g\left( x \right)} \right)}^2}}}\]
Now, substituting \[g(x) = \dfrac{{2x}}{{1 - {x^2}}}\] in above expression, we have
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{1}{{1 + {{\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)}^2}}}\]
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{1}{{1 + \dfrac{{{{\left( {2x} \right)}^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}}}\]
Now, taking LCM in the denominator.
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{1}{{\dfrac{{{{\left( {1 - {x^2}} \right)}^2} + {{\left( {2x} \right)}^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}}}\]
Using the property \[\dfrac{1}{{\dfrac{a}{b}}} = \dfrac{b}{a}\], we get
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{{{{\left( {1 - {x^2}} \right)}^2}}}{{{{\left( {1 - {x^2}} \right)}^2} + {{\left( {2x} \right)}^2}}}\]
Now, opening the brackets using the property \[{\left( {a - b} \right)^2} = {\left( a \right)^2} + {\left( b \right)^2} - 2ab\]
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{{{{\left( 1 \right)}^2} + {{\left( {{x^2}} \right)}^2} - \left( {2 \times 1 \times {x^2}} \right)}}{{\left( {{{\left( 1 \right)}^2} + {{\left( {{x^2}} \right)}^2} - \left( {2 \times 1 \times {x^2}} \right)} \right) + {{\left( {2x} \right)}^2}}}\]
Using \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], we have
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{{1 + {{\left( x \right)}^{2 \times 2}} - 2{x^2}}}{{\left( {1 + {{\left( x \right)}^{2 \times 2}} - 2{x^2}} \right) + 4{x^2}}}\]
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{{1 + {x^4} - 2{x^2}}}{{\left( {1 + {x^4} - 2{x^2}} \right) + 4{x^2}}}\]
Opening the brackets, we have
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{{1 + {x^4} - 2{x^2}}}{{1 + {x^4} - 2{x^2} + 4{x^2}}}\]
Solving the like terms in denominator, we get
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{{1 + {x^4} - 2{x^2}}}{{1 + {x^4} + 2{x^2}}}\]
We will use the properties \[\left( {{{(a + b)}^2} = {a^2} + {b^2} + 2ab} \right)\] and \[\left( {{{(a - b)}^2} = {a^2} + {b^2} - 2ab} \right)\] by taking \[a = 1\] and \[b = {x^2}\] we have,
 \[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{{{{\left( 1 \right)}^2} + {{\left( {{x^2}} \right)}^2} - 2 \times 1 \times {x^2}}}{{{{\left( 1 \right)}^2} + {{\left( {{x^2}} \right)}^2} + 2 \times 1 \times {x^2}}}\]
\[ \Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{{{{\left( {1 - {x^2}} \right)}^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} - - - - - - (5)\]
Now, finding \[g'(x)\]
From (2), we have
\[\dfrac{{dg}}{{dx}} = \dfrac{d}{{dx}}\left( {g(x)} \right)\]
\[ \Rightarrow \dfrac{{dg}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)\]
Since the given expression is a fraction, we will apply the quotient rule.
Let \[u = 2x - - - - - - (6)\]
And \[v = 1 - {x^2} - - - - - - (7)\]
Now, using (6), (7) and the quotient rule i.e. \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}}\], we have
\[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}}\], where \[u' = \dfrac{d}{{dx}}\left( {u\left( x \right)} \right)\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{\left( {\left( {1 - {x^2}} \right) \times \dfrac{d}{{dx}}\left( {2x} \right)} \right) - \left( {\left( {2x} \right) \times \dfrac{d}{{dx}}\left( {1 - {x^2}} \right)} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
Using the properties \[\dfrac{d}{{dx}}\left( {cf(x)} \right) = c \times \dfrac{d}{{dx}}\left( {f(x)} \right)\], where \[c\] is a constant function and \[\dfrac{d}{{dx}}\left( {f(x) - g(x)} \right) = \dfrac{d}{{dx}}\left( {f(x)} \right) - \dfrac{d}{{dx}}\left( {g(x)} \right)\], we have
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{\left( {\left( {1 - {x^2}} \right) \times \left( {2 \times \dfrac{d}{{dx}}\left( x \right)} \right)} \right) - \left( {\left( {2x} \right) \times \left( {\dfrac{d}{{dx}}\left( 1 \right) - \dfrac{d}{{dx}}\left( {{x^2}} \right)} \right)} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
Now, using formulae \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] and \[\dfrac{d}{{dx}}\left( c \right) = 0\], where \[c\] is a constant.
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{\left( {\left( {1 - {x^2}} \right) \times \left( {2 \times \left( {1{x^{1 - 1}}} \right)} \right)} \right) - \left( {\left( {2x} \right) \times \left( {0 - \left( {2{x^{2 - 1}}} \right)} \right)} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{\left( {\left( {1 - {x^2}} \right) \times \left( {2 \times \left( {{x^0}} \right)} \right)} \right) - \left( {\left( {2x} \right) \times \left( { - 2{x^1}} \right)} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
Now, using \[{x^0} = 1\] and \[{x^1} = x\], we have
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{\left( {\left( {1 - {x^2}} \right) \times \left( {2 \times \left( 1 \right)} \right)} \right) - \left( {\left( {2x} \right) \times \left( { - 2x} \right)} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
Solving the brackets, we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{\left( {\left( {1 - {x^2}} \right) \times \left( 2 \right)} \right) - \left( {\left( {2x} \right) \times \left( { - 2x} \right)} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{\left( {2\left( {1 - {x^2}} \right)} \right) - \left( { - 4{x^2}} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
Using \[ - \left( { - f(x)} \right) = + f(x)\] and opening the brackets, we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{2 - 2{x^2} + 4{x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
Now, solving the like terms in the numerator, we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{2 + 2{x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
\[ \Rightarrow g'(x) = \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{2 + 2{x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
Taking out \[2\] common from the numerator, we get
\[ \Rightarrow g'(x) = \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{2\left( {1 + {x^2}} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}} - - - - - - (8)\]
Using (4), (5) and (8), we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {fog(x)} \right) = \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)} \right) = f'\left( {g\left( x \right)} \right) \times g'(x)\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)} \right) = \dfrac{{{{\left( {1 - {x^2}} \right)}^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \times \dfrac{{2\left( {1 + {x^2}} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}\]
Now, cancelling out the terms from numerator and denominator, we have
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)} \right) = \dfrac{1}{{\left( {1 + {x^2}} \right)}} \times \dfrac{2}{1}\]
Multiplying the numerator and denominator of both the terms, we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)} \right) = \dfrac{2}{{\left( {1 + {x^2}} \right)}}\]
Hence, derivative of \[{\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)\] is \[\dfrac{2}{{\left( {1 + {x^2}} \right)}}\].

Note: In this question, we should take care that we have to apply two rules. Whenever we see the expression as a composition of two functions, we need to use chain rule to find its derivative and to find the derivative of any fraction terms, we have to use quotient rule. While choosing the functions to form \[fog(x)\], we have to be careful which function should be chosen as \[f(x)\] and \[g(x)\]. When we are calculating the terms, we should go step by step in order to avoid mistakes. Also, we need to revise all the identities and use them both ways.