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

If we have a function as y=tan⁡−1(2x1−x2)y = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right), then find dydx\dfrac{{dy}}{{dx}}.

Explanation

Solution

We need to find the derivative of tan⁡−1(2x1−x2){\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right). We see that this function is the composition of two functions i.e. tan⁡−1(x){\tan ^{ - 1}}\left( x \right) and 2x1−x2\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
ddx(fog(x))=f′(g(x))×g′(x)\dfrac{d}{{dx}}\left( {fog(x)} \right) = f'(g(x)) \times g'(x), where g′(x)=dgdxg'(x) = \dfrac{{dg}}{{dx}}.
Here, we again see that 2x1−x2\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:
ddx(uv)=vu′−uv′v2\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}}, where uu and vv are functions of xx and v′=dvdxv' = \dfrac{{dv}}{{dx}}

Complete step-by-step solution:
Finding the derivative of tan⁡−1(2x1−x2){\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right). Now, writing this function as a composition of two functions.
Let f(x)=tan⁡−1x−−−−−−(1)f(x) = {\tan ^{ - 1}}x - - - - - - (1)
And g(x)=2x1−x2−−−−−−(2)g(x) = \dfrac{{2x}}{{1 - {x^2}}} - - - - - - (2)
Using (1) and (2), we get
fog(x)=f(g(x))fog(x) = f\left( {g\left( x \right)} \right)
⇒fog(x)=tan⁡−1(g(x))\Rightarrow fog(x) = {\tan ^{ - 1}}\left( {g\left( x \right)} \right)
Substituting g(x)=2x1−x2g(x) = \dfrac{{2x}}{{1 - {x^2}}} in above expression, we have
⇒fog(x)=tan⁡−1(2x1−x2)−−−−−−(3)\Rightarrow fog(x) = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) - - - - - - (3)
Now, to find the derivative of this function, we have
ddx(fog(x))=f′(g(x))×g′(x)\dfrac{d}{{dx}}\left( {fog(x)} \right) = f'(g(x)) \times g'(x)
⇒ddx(fog(x))=ddx(tan⁡−1(2x1−x2))=f′(g(x))×g′(x)−−−−−−(4)\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′(g(x))f'\left( {g\left( x \right)} \right) first
For that we first have to find f′(x)=dfdxf'\left( x \right) = \dfrac{{df}}{{dx}}
From (1), we have
dfdx=ddx(tan⁡−1(x))\dfrac{{df}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( x \right)} \right)
As we know, ddx(tan⁡−1(x))=11+x2\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( x \right)} \right) = \dfrac{1}{{1 + {x^2}}}. Using in the above formula, we have
⇒dfdx=ddx(tan⁡−1(x))=11+x2\Rightarrow \dfrac{{df}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( x \right)} \right) = \dfrac{1}{{1 + {x^2}}}
⇒dfdx=f′(x)=11+x2\Rightarrow \dfrac{{df}}{{dx}} = f'\left( x \right) = \dfrac{1}{{1 + {x^2}}}
Now, replacing xx by g(x)g(x), we have
⇒f′(g(x))=11+(g(x))2\Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{1}{{1 + {{\left( {g\left( x \right)} \right)}^2}}}
Now, substituting g(x)=2x1−x2g(x) = \dfrac{{2x}}{{1 - {x^2}}} in above expression, we have
⇒f′(g(x))=11+(2x1−x2)2\Rightarrow f'\left( {g\left( x \right)} \right) = \dfrac{1}{{1 + {{\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)}^2}}}
⇒f′(g(x))=11+(2x)2(1−x2)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.
⇒f′(g(x))=1(1−x2)2+(2x)2(1−x2)2\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 1ab=ba\dfrac{1}{{\dfrac{a}{b}}} = \dfrac{b}{a}, we get
⇒f′(g(x))=(1−x2)2(1−x2)2+(2x)2\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 (a−b)2=(a)2+(b)2−2ab{\left( {a - b} \right)^2} = {\left( a \right)^2} + {\left( b \right)^2} - 2ab
⇒f′(g(x))=(1)2+(x2)2−(2×1×x2)((1)2+(x2)2−(2×1×x2))+(2x)2\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 (am)n=amn{\left( {{a^m}} \right)^n} = {a^{mn}}, we have
⇒f′(g(x))=1+(x)2×2−2x2(1+(x)2×2−2x2)+4x2\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}}}
⇒f′(g(x))=1+x4−2x2(1+x4−2x2)+4x2\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
⇒f′(g(x))=1+x4−2x21+x4−2x2+4x2\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
⇒f′(g(x))=1+x4−2x21+x4+2x2\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 ((a+b)2=a2+b2+2ab)\left( {{{(a + b)}^2} = {a^2} + {b^2} + 2ab} \right) and ((a−b)2=a2+b2−2ab)\left( {{{(a - b)}^2} = {a^2} + {b^2} - 2ab} \right) by taking a=1a = 1 and b=x2b = {x^2} we have,
⇒f′(g(x))=(1)2+(x2)2−2×1×x2(1)2+(x2)2+2×1×x2\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}}}
⇒f′(g(x))=(1−x2)2(1+x2)2−−−−−−(5)\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)g'(x)
From (2), we have
dgdx=ddx(g(x))\dfrac{{dg}}{{dx}} = \dfrac{d}{{dx}}\left( {g(x)} \right)
⇒dgdx=ddx(2x1−x2)\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)u = 2x - - - - - - (6)
And v=1−x2−−−−−−(7)v = 1 - {x^2} - - - - - - (7)
Now, using (6), (7) and the quotient rule i.e. ddx(uv)=vu′−uv′v2\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}}, we have
ddx(uv)=ddx(2x1−x2)=vu′−uv′v2\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′=ddx(u(x))u' = \dfrac{d}{{dx}}\left( {u\left( x \right)} \right)
⇒ddx(2x1−x2)=((1−x2)×ddx(2x))−((2x)×ddx(1−x2))(1−x2)2\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 ddx(cf(x))=c×ddx(f(x))\dfrac{d}{{dx}}\left( {cf(x)} \right) = c \times \dfrac{d}{{dx}}\left( {f(x)} \right), where cc is a constant function and ddx(f(x)−g(x))=ddx(f(x))−ddx(g(x))\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
⇒ddx(2x1−x2)=((1−x2)×(2×ddx(x)))−((2x)×(ddx(1)−ddx(x2)))(1−x2)2\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 ddx(xn)=nxn−1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} and ddx(c)=0\dfrac{d}{{dx}}\left( c \right) = 0, where cc is a constant.
⇒ddx(2x1−x2)=((1−x2)×(2×(1x1−1)))−((2x)×(0−(2x2−1)))(1−x2)2\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}}}
⇒ddx(2x1−x2)=((1−x2)×(2×(x0)))−((2x)×(−2x1))(1−x2)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 x0=1{x^0} = 1 and x1=x{x^1} = x, we have
⇒ddx(2x1−x2)=((1−x2)×(2×(1)))−((2x)×(−2x))(1−x2)2\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
⇒ddx(2x1−x2)=((1−x2)×(2))−((2x)×(−2x))(1−x2)2\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}}}
⇒ddx(2x1−x2)=(2(1−x2))−(−4x2)(1−x2)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 −(−f(x))=+f(x) - \left( { - f(x)} \right) = + f(x) and opening the brackets, we get
⇒ddx(2x1−x2)=2−2x2+4x2(1−x2)2\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
⇒ddx(2x1−x2)=2+2x2(1−x2)2\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) = \dfrac{{2 + 2{x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}
⇒g′(x)=ddx(2x1−x2)=2+2x2(1−x2)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 22 common from the numerator, we get
⇒g′(x)=ddx(2x1−x2)=2(1+x2)(1−x2)2−−−−−−(8)\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
⇒ddx(fog(x))=ddx(tan⁡−1(2x1−x2))=f′(g(x))×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)
⇒ddx(tan⁡−1(2x1−x2))=(1−x2)2(1+x2)2×2(1+x2)(1−x2)2\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
⇒ddx(tan⁡−1(2x1−x2))=1(1+x2)×21\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
⇒ddx(tan⁡−1(2x1−x2))=2(1+x2)\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(2x1−x2){\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) is 2(1+x2)\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)fog(x), we have to be careful which function should be chosen as f(x)f(x) and g(x)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.