Question

The derivative of ${\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ with respect to ${\tan ^{ - 1}}x$ is

Answer 1

Hint : First we know a differential equation is the equation which contains dependent variables, independent variables and derivatives of the dependent variables with respect to the independent variables. Since differentiation of a given dependent function with respect to the independent variable is a function of another independent variable. So, we have to use chain rule to find its derivative

** Complete step-by-step answer** :
Complete step by step solution: Differential equations are classified into two types, Ordinary differential equations where dependent variables depend on only one independent variable and Partial differential equations where dependent variables depend on two or more independent variables.
Suppose $u$ and $v$ are functions of $x$ only. Then differentiation has the following rules:
1.The derivative of the constant function is equal to zero.
2.Product rule: $\dfrac{d}{{dx}}\left( {u \times v} \right) = v \times \dfrac{d}{{dx}}\left( u \right) + u \times \dfrac{d}{{dx}}\left( v \right)$ .
3.Quotient rule: $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v \times \dfrac{d}{{dx}}\left( u \right) - u \times \dfrac{d}{{dx}}\left( v \right)}}{{{v^2}}}$ .
4.If $u$ is a function of $v$ , Then $\dfrac{d}{{dx}}\left( {u(v)} \right) = \dfrac{d}{{dv}}\left( {u(v)} \right).\dfrac{{dv}}{{dx}}$
Given $u = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ ---(1) and $v = {\tan ^{ - 1}}\left( x \right)$ .---(2)
Here we have to find the derivative of $u$ with respect to $v$ where $u$ and $v$ is a function of $x$ .
Differentiating with respect to $x$ both sides of the equation (1), we get
$\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{\sqrt {1 + {x^2}} - 1}}{x}} \right) = \dfrac{{\left( {\dfrac{x}{{\sqrt {1 + {x^2}} }} - 0} \right) \times x - \left( {\sqrt {1 + {x^2}} - 1} \right)}}{{{x^2}}}$
Simplifying the above equation, we get
$\dfrac{{du}}{{dx}} = \dfrac{{{x^2} - 1 - {x^2} + \sqrt {1 + {x^2}} }}{{{x^2}\sqrt {1 + {x^2}} }} = \dfrac{{\sqrt {1 + {x^2}} - 1}}{{{x^2}\sqrt {1 + {x^2}} }}$ ----(3)
Differentiating with respect to $x$ both sides of the equation (2), we get
$\dfrac{{dv}}{{dx}} = \dfrac{1}{{1 + {x^2}}}$ -----(4)
Then form 4th rule we get, $\dfrac{{du}}{{dv}} = \dfrac{{du}}{{dx}} \times \dfrac{{dx}}{{dv}} = \dfrac{{du}}{{dx}} \times \dfrac{1}{{\dfrac{{dv}}{{dx}}}}$
From the equations (3) and (4), we get
$\dfrac{{du}}{{dv}} = \dfrac{{\sqrt {1 + {x^2}} - 1}}{{{x^2}\sqrt {1 + {x^2}} }} \times \dfrac{1}{{\dfrac{1}{{1 + {x^2}}}}} = \dfrac{{\sqrt {1 + {x^2}} \left( {\sqrt {1 + {x^2}} - 1} \right)}}{{{x^2}}}$ .
Hence, the derivative of ${\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ with respect to ${\tan ^{ - 1}}x$ is $\dfrac{{\sqrt {1 + {x^2}} \left( {\sqrt {1 + {x^2}} - 1} \right)}}{{{x^2}}}$ .
So, the correct answer is “ $\dfrac{{\sqrt {1 + {x^2}} \left( {\sqrt {1 + {x^2}} - 1} \right)}}{{{x^2}}}$ .”.

Note : Note that ${4^{th}}$ rule is known as the chain rule. If the function is composed of three functions, say $u$ , $v$ and $w$ are functions of $x$ . Then the derivative of the composition of three function is defined as follows $\dfrac{d}{{dx}}\left( {u(v(w(x)))} \right) = \dfrac{{du}}{{dv}}.\dfrac{{dv}}{{dw}}.\dfrac{{dw}}{{dx}}$ .