Question

The value of $\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left\{ {\dfrac{{\sqrt x \left( {3 - x} \right)}}{{1 - 3x}}} \right\}} \right]$
a) $\dfrac{1}{{2\left( {1 + x} \right)\sqrt x }}$
b)$\dfrac{3}{{\left( {1 + x} \right)\sqrt x }}$
c) $\dfrac{2}{{\left( {1 + x} \right)\sqrt x }}$
d)$\dfrac{3}{{2\left( {1 + x} \right)\sqrt x }}$

Answer 1

Hint: Here we are asked to find the derivative of the given expression. We will solve this problem by using a substitution method. Substitution method can be used to convert the complex terms into simpler forms which will help us to find the derivative easily. After finding the derivative of the substituted function we have to re-substitute the terms again to represent the result in original variables as given in the question.

Formula: Formula that we need to know:
$\dfrac{d}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }}$
$\dfrac{d}{{dx}}{\tan ^{ - 1}} = \dfrac{1}{{1 + {x^2}}}$
$\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$

Complete step-by-step answer:
We aim to find the derivative of the expression $\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left\{ {\dfrac{{\sqrt x \left( {3 - x} \right)}}{{1 - 3x}}} \right\}} \right]$ for this we first need to make some substitution to make this complex expression into a simpler expression.
Let $y = {\tan ^{ - 1}}\left\{ {\dfrac{{\sqrt x \left( {3 - x} \right)}}{{1 - 3x}}} \right\}$ we aim to find the value of $\dfrac{{dy}}{{dx}}$ .
Let us substitute $x = {\tan ^2}\theta $ in the given expression.
$y = {\tan ^{ - 1}}\left\{ {\dfrac{{\sqrt {{{\tan }^2}\theta } \left( {3 - {{\tan }^2}\theta } \right)}}{{1 - 3{{\tan }^2}\theta }}} \right\}$
On simplifying the above, we get
$y = {\tan ^{ - 1}}\left\{ {\dfrac{{\tan \theta \left( {3 - {{\tan }^2}\theta } \right)}}{{1 - 3{{\tan }^2}\theta }}} \right\}$
On further simplification we get
$y = {\tan ^{ - 1}}\left\{ {\dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right\}$
Now using the formula $\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$ we get
$y = {\tan ^{ - 1}}\left\{ {\tan 3\theta } \right\}$
$y = 3\theta $
Now we have to re-substitute the value that we substituted before. That is $x = {\tan ^2}\theta $
From this we have to find the value of theta to re-substitute it.
$x = {\tan ^2}\theta $
$\tan \theta = \sqrt x $
$\theta = {\tan ^{ - 1}}\sqrt x $
Thus, we got the value of theta let’s substitute it in the equation $y = 3\theta $
$y = 3{\tan ^{ - 1}}\sqrt x $
Now let us find the derivative oof the above function.
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {3{{\tan }^{ - 1}}\sqrt x } \right)$
On differentiating the above with respect to $x$ using the formula $\dfrac{d}{{dx}}{\tan ^{ - 1}} = \dfrac{1}{{1 + {x^2}}}$ and $\dfrac{d}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }}$ we get
$\dfrac{{dy}}{{dx}} = 3 \times \left( {\dfrac{1}{{\left( {1 + {{\sqrt x }^2}} \right)2\sqrt x }}} \right)$
On simplifying the above, we get
$\dfrac{{dy}}{{dx}} = \dfrac{3}{{2\sqrt x \left( {1 + x} \right)}}$
$\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left\{ {\dfrac{{\sqrt x \left( {3 + x} \right)}}{{1 - 3x}}} \right\}} \right] = \dfrac{3}{{2\sqrt x \left( {1 + x} \right)}}$
Thus, we have found the value of the derivative of the given expression.
Now let us find the correct option from the given.
Option (a) $\dfrac{1}{{2\left( {1 + x} \right)\sqrt x }}$ is not the correct answer as we got the value as $\dfrac{3}{{2\left( {1 + x} \right)\sqrt x }}$ from our calculation above.
Option (b) $\dfrac{3}{{\left( {1 + x} \right)\sqrt x }}$ is not the correct answer as we got the value as $\dfrac{3}{{2\left( {1 + x} \right)\sqrt x }}$ from our calculation above.
Option (c) $\dfrac{2}{{\left( {1 + x} \right)\sqrt x }}$ is not the correct answer as we got the value as $\dfrac{3}{{2\left( {1 + x} \right)\sqrt x }}$ from our calculation above.
Option (d) $\dfrac{3}{{2\left( {1 + x} \right)\sqrt x }}$ is the correct answer as we got the same value in our calculation above.
Hence, option (d) $\dfrac{3}{{2\left( {1 + x} \right)\sqrt x }}$ is the correct option.

So, the correct answer is “Option (d)”.

Note: The derivative of function is nothing but the rate of change in that function. The derivative also has orders like first-order derivative, second-order derivative, third order-derivative, and so on. The derivative of a constant is zero. So, the order of the derivative stops when we attain zero for any derivative.