Question

Differentiate ${\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ with respect to ${\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$ , when $x \ne 0$.

Answer 1

Hint : In this differentiation problem we first let one function as ‘u’ and other as ‘v’ and then simplify the angle of inverse trigonometric functions given by using suitable substitution. After simplification of both, divide their differentiation result to obtain the required solution of the problem.

Formula used:

${\sec ^2}\theta - {\tan ^2}\theta = 1,\,\,\,\,\,$

$\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }},\,\,$

$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }},\,\,\,$

$\sec \theta = \dfrac{1}{{\cos \theta }},\,\,$

$\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}},\,\,1 - \cos \theta = 2{\sin ^2}\left( {\dfrac{\theta}{2}} \right)\,\,\,$

\sin \theta = 2.\sin \left( {\dfrac{\theta}{2}} \right)\\\cos \left( {\dfrac{\theta}{2}} \right)\

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Complete step by step solution:**

Suppose

$u = \dfrac{{{{\tan }^{ - 1}}\sqrt {1 + {x^2}} - 1}}{x}$

We know that for an inverse trigonometric function having $\sqrt {1 + {x^2}}$ term suitable substitution is $x = \tan \theta$ .

Therefore taking $x = \tan \theta$ in above we have,

$u = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}} \right)$

$\Rightarrow u = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt {{{\sec }^2}\theta } - 1}}{{\tan \theta }}} \right)$

\,\,\,\,\,\,\,\,\,\,\,\,\left\\{ {\because 1 + {{\tan }^2}\theta = {{\sec }^2}\theta } \right\\}

$\Rightarrow u = {\tan ^{ - 1}}\left( {\dfrac{{\sec \theta - 1}}{{\tan \theta }}} \right)$

Writing $\sec \theta \,\,and\,\,\tan \theta \,\,\operatorname{int} o\,\,basic\,\,trigonometric\,\,functions$

u = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{{\cos \theta }} - 1}}{{\dfrac{{\sin \theta }}{{\cos \theta }}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \left\\{ {\sec \theta = \dfrac{1}{{\cos \theta }}and\,\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}} \right\\}

Taking L.C.M. we have,

$u = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{1 - \cos \theta }}{{\cos \theta }}}}{{\dfrac{{\sin \theta }}{{\cos \theta }}}}} \right)$

or

$u = {\tan ^{ - 1}}\left( {\dfrac{{1 - \cos \theta }}{{\sin \theta }}} \right)$

Now, using half angle of identities of trigonometric functions $1 - \cos \theta = 2{\sin ^2}\left( \dfrac{\theta }{2} \right),\,\,and\,\,\sin \theta = 2.\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)$ we have

$u = {\tan ^{ - 1}}\left( {\dfrac{{2{{\sin }^2}\left( \dfrac{\theta }{2} \right)}}{{2.\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)}}} \right)$

$u = {\tan ^{ - 1}}\left( {\dfrac{{\sin \left( \dfrac{\theta }{2} \right)}}{{\cos \left( \dfrac{\theta }{2} \right)}}} \right)$

u = {\tan ^{ - 1}}\left\\{ {\tan \left( \dfrac{\theta }{2} \right)} \right\\}

u = \dfrac{\theta }{2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\\{ {\because {{\tan }^{ - 1}}\left( {\tan \theta } \right) = \theta } \right\\}

From substitution $x = \tan \theta \,\,\,we\,\,have\,\,\theta = {\tan ^{ - 1}}x$ . Using this in above we have

$u = \dfrac{{{{\tan }^{ - 1}}x}}{2}$ or

$u = \dfrac{1}{2}{\tan ^{ - 1}}x$

Now, differentiating ‘u’ w.r.t. to x we have

$\dfrac{{du}}{{dx}} = \dfrac{1}{2}\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right)$

$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{2}\left( {\dfrac{1}{{1 + {x^2}}}} \right)$

\left\\{ {\because \dfrac{d}{{dx}}{{\tan }^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}} \right\\}

Or

$\dfrac{{du}}{{dx}} = \dfrac{1}{2}\left( {\dfrac{1}{{1 + {x^2}}}} \right).......................(i)$

Now, considering v = ${\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$

Taking, x = $\tan \theta$ in above and simplifying it we have,

$v = {\sin ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)$

\Rightarrow v = {\sin ^{ - 1}}\left( {\sin 2\theta } \right) ,\,\,\,\,\,\,\,\,\,\,\,\left\\{ {\because \sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right\\}

v = 2\theta ,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\\{ {\because {{\sin }^{ - 1}}\left( {\sin \theta } \right) = \theta } \right\\}

From substitution $x = \tan \theta \,\,we\,\,have\,\,\theta = {\tan ^{ - 1}}x$ . Using this in above we have

$v = 2{\tan ^{ - 1}}x$

Now, differentiating ‘v’ w.r.t. x we have,

$\dfrac{{dv}}{{dx}} = \dfrac{d}{{d\theta }}(2{\tan ^{ - 1}}x)$

$\Rightarrow \dfrac{{dv}}{{dx}} = 2\dfrac{d}{{d\theta }}({\tan ^{ - 1}}x)$

\Rightarrow \dfrac{{dv}}{{dx}} = 2\left( {\dfrac{1}{{1 + {x^2}}}} \right),\,\,\,\,\,\,\,\,\left\\{ {{{\tan }^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}} \right\\}

or

$\dfrac{{dv}}{{dx}} = \dfrac{2}{{1 + {x^2}}}........................(ii)$

Now, dividing equation (i) by equation (ii) we have

$\dfrac{{\left( {\dfrac{{du}}{{d\theta }}} \right)}}{{\left( {\dfrac{{dv}}{{d\theta }}} \right)}} = \dfrac{{\dfrac{1}{2}\left( {\dfrac{1}{{1 + {x^2}}}} \right)}}{{2\left( {\dfrac{1}{{1 + {x^2}}}} \right)}}$

$\Rightarrow \dfrac{{du}}{{dv}} = \dfrac{1}{4}$

Therefore, from above we see that derivative of ‘u’ w.r.t. ‘v’ is $\dfrac{1}{4}$ .

Hence, we can say that derivative of ${\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ w.r.t. ${\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$ is $\dfrac{1}{4}$ .

So, the correct answer is “ $\dfrac{1}{4}$ ”.

Note : For inverses trigonometric functions we first see that if angles of the given functions are not in simplified form then we must use suitable substitution to convert given angle of trigonometric function in lowest form of simplified form and only then differentiate them otherwise calculating ends with wrong answer.