Question

Differentiate the following functions with respect to $x$
(1) \sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \right]} \right\\}
(2) ${\cot ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} }}{x}} \right)$
(3) ${\tan ^{ - 1}}\left( {\dfrac{{a + x}}{{1 - ax}}} \right)$

Answer 1

We use many differentiation formulas and differentiating techniques to solve this problem. We will use substitution methods and some simplification methods and solve this problem. In the substitution method, we will substitute a value for a variable, which makes our problem simpler.

Complete answer:
(1)
We have to differentiate \sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \right]} \right\\} with respect to $x$.
So, we have to find the value of \dfrac{d}{{dx}}\sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \right]} \right\\}
So, we will use a substitution method to solve this problem.
We will substitute $x = \cos 2\theta$ ------(a)
So, we get it as
\dfrac{d}{{dx}}\sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \right]} \right\\} = \dfrac{d}{{dx}}\sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\sqrt {\dfrac{{1 - \cos 2\theta }}{{1 + \cos 2\theta }}} } \right]} \right\\}
We all know the formula $\dfrac{{1 - \cos 2\theta }}{{1 + \cos 2\theta }} = {\tan ^2}\theta$
So, we can write it as
\Rightarrow \dfrac{d}{{dx}}\sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \right]} \right\\} = \dfrac{d}{{dx}}\sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\sqrt {{{\tan }^2}\theta } } \right]} \right\\}
\Rightarrow \dfrac{d}{{dx}}\sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \right]} \right\\} = \dfrac{d}{{dx}}\sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\tan \theta } \right]} \right\\}
As we know that, ${\tan ^{ - 1}}(\tan \theta ) = \theta$
The equation will change as,
\Rightarrow \dfrac{d}{{dx}}\sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \right]} \right\\} = \dfrac{d}{{dx}}\sin \left\\{ {2\theta } \right\\}
So, now on differentiating this, we get,
\Rightarrow \dfrac{d}{{dx}}\sin \left\\{ {2\theta } \right\\} = 2\cos 2\theta .\dfrac{{d\theta }}{{dx}} $\left( {\because \dfrac{d}{{dx}}\sin x = \cos x} \right)$ and $\left( {\because \dfrac{d}{{dx}}\cos x = - \sin x} \right)$
From equation (a), $x = \cos 2\theta$, differentiating this with respect to $x$, we get, $1 = - 2\sin 2\theta .\dfrac{{d\theta }}{{dx}}$
So, from this, we can write, $\dfrac{{d\theta }}{{dx}} = \dfrac{{ - 1}}{{2\sin 2\theta }}$
Now substituting this value in our main result, we get,
\dfrac{d}{{dx}}\sin \left\\{ {2\theta } \right\\} = - 2\cos 2\theta .\dfrac{{ - 1}}{{2\sin 2\theta }}
\Rightarrow \dfrac{d}{{dx}}\sin \left\\{ {2\theta } \right\\} = \cot 2\theta
Now, substituting back the real values in terms of $x$, we get the final result as,
\dfrac{d}{{dx}}\sin \left\\{ {2{{\tan }^{ - 1}}\left[ {\sqrt {\dfrac{{1 - x}}{{1 + x}}} } \right]} \right\\} = \cot \left[ {{{\cos }^{ - 1}}x} \right]
This is the required answer.

(2)
We have to differentiate ${\cot ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} }}{x}} \right)$ with respect to $x$.
So, we have to find $\dfrac{d}{{dx}}{\cot ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} }}{x}} \right)$
So, we will follow chain rule to solve this problem.
Chain rule is defined as $\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)$
$\Rightarrow \dfrac{d}{{dx}}{\cot ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} }}{x}} \right) = \dfrac{{ - 1}}{{1 + {{\left( {\dfrac{{\sqrt {1 + {x^2}} }}{x}} \right)}^2}}}\dfrac{d}{{dx}}\left( {\dfrac{{\sqrt {1 + {x^2}} }}{x}} \right)$ $\left( {\because \dfrac{d}{{dt}}{{\cot }^{ - 1}}t = \dfrac{{ - 1}}{{1 + {t^2}}}} \right)$

$\Rightarrow \dfrac{{ - 1}}{{1 + \left( {\dfrac{{1 + {x^2}}}{{{x^2}}}} \right)}}\left( {\dfrac{{x\dfrac{d}{{dx}}\sqrt {1 + {x^2}} - \sqrt {1 + {x^2}} \dfrac{d}{{dx}}x}}{{{x^2}}}} \right)$ $\left( {\because \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v.du - u.dv}}{{{v^2}}}} \right)$
Now, we have to simplify this as follows.
$\Rightarrow \dfrac{{ - 1}}{{\dfrac{{{x^2} + 1 + {x^2}}}{{{x^2}}}}}\left( {\dfrac{{x\left( {\dfrac{1}{{2\sqrt {1 + {x^2}} }}\dfrac{d}{{dx}}(1 + {x^2})} \right) - \sqrt {1 + {x^2}} }}{{{x^2}}}} \right)$ $\left( {\because \dfrac{d}{{dt}}\sqrt t = \dfrac{1}{{2\sqrt t }}} \right)$ and $\left( {\because \dfrac{d}{{dt}}{t^n} = n{t^{n - 1}}} \right)$
$\Rightarrow \dfrac{{ - {x^2}}}{{2{x^2} + 1}}\left( {\dfrac{{x\left( {\dfrac{{2x}}{{2\sqrt {1 + {x^2}} }}} \right) - \sqrt {1 + {x^2}} }}{{{x^2}}}} \right)$
So, finally we get the result as,
$\Rightarrow \dfrac{{ - 1}}{{2{x^2} + 1}}\left( {\left( {\dfrac{{{x^2}}}{{\sqrt {1 + {x^2}} }}} \right) - \sqrt {1 + {x^2}} } \right)$
$\Rightarrow \dfrac{{ - 1}}{{2{x^2} + 1}}\left( {\dfrac{{{x^2} - (1 + {x^2})}}{{\sqrt {1 + {x^2}} }}} \right) = \dfrac{1}{{(2{x^2} + 1)\sqrt {1 + {x^2}} }}$
So, finally, we can conclude that,
$\dfrac{d}{{dx}}{\cot ^{ - 1}}\left( {\dfrac{{\sqrt {1 + {x^2}} }}{x}} \right) = \dfrac{1}{{(2{x^2} + 1)\sqrt {1 + {x^2}} }}$
This is the required result.

(3)
We have to differentiate ${\tan ^{ - 1}}\left( {\dfrac{{a + x}}{{1 - ax}}} \right)$ with respect to $x$
So, we have to find $\dfrac{d}{{dx}}{\tan ^{ - 1}}\left( {\dfrac{{a + x}}{{1 - ax}}} \right)$
Take $a = \tan m$ and $x = \tan n$. So, when you differentiate the second substitution, we get, $1 = {\sec ^2}n.\dfrac{{dn}}{{dx}}$ $\left( {\because \dfrac{d}{{dt}}\tan t = {{\sec }^2}t} \right)$
So, on substitution, we get,
$\dfrac{d}{{dx}}{\tan ^{ - 1}}\left( {\dfrac{{a + x}}{{1 - ax}}} \right) = \dfrac{d}{{dx}}{\tan ^{ - 1}}\left( {\dfrac{{\tan m + \tan n}}{{1 - \tan m.\tan n}}} \right)$
We know that, $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}$
$\Rightarrow \dfrac{d}{{dx}}{\tan ^{ - 1}}\left( {\tan (m + n)} \right)$
And we know that, ${\tan ^{ - 1}}(\tan \theta ) = \theta$
So, we get,
$\Rightarrow \dfrac{d}{{dx}}\left( {m + n} \right) = \dfrac{d}{{dx}}m + \dfrac{d}{{dx}}n$
As, $m$ is constant,
V
$\Rightarrow \dfrac{d}{{dx}}\left( {m + n} \right) = 0 + \dfrac{1}{{{{\sec }^2}n}}$
We know that, ${\sec ^2}\theta - {\tan ^2}\theta = 1$
From this, we can write as,
$\Rightarrow \dfrac{1}{{1 + {{\tan }^2}n}} = \dfrac{1}{{1 + {x^2}}}$
So, finally the result is
$\dfrac{d}{{dx}}{\tan ^{ - 1}}\left( {\dfrac{{a + x}}{{1 - ax}}} \right) = \dfrac{1}{{1 + {x^2}}}$
This is the required result.

Note: Differentiation of a constant is always 0. And also make a note that, $\dfrac{d}{{dx}}f(y) = f'(y).\dfrac{{dy}}{{dx}}$.
Also remember the formula ${\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) = {\tan ^{ - 1}}x + {\tan ^{ - 1}}y$ which will be very useful to you.
Differentiation of a function of another variable is always zero i.e., $\dfrac{d}{{dx}}f(y) = 0$ if $x{\text{ and }}y$ are not related.