Question

If for $x \in \left( {0,\dfrac{1}{4}} \right),$ the derivative of ${\tan ^{ - 1}}\left( {\dfrac{{6x\sqrt x }}{{1 - 9{x^3}}}} \right)$ is $\sqrt x .g\left( x \right),$ then $g\left( x \right)$ equals.
A. $\dfrac{9}{{1 + 9{x^3}}}$
B. $\dfrac{{3x}}{{1 - 9{x^3}}}$
C. $\dfrac{{3x\sqrt x }}{{1 - 9{x^3}}}$
D. $\dfrac{3}{{1 + 9{x^3}}}$

Answer 1

Differentiation is the action of completing a derivation, the derivation of a function $y = f\left( x \right)$ of a variable x is a measure of the rate at which the value of the function changes with respect to the change of the variable x, and it is called the derivative of f with respect to x. In this question $2{\tan ^{ - 1}}x{\tan ^{ - 1}}\left( {\dfrac{{2 \times x}}{{1 - {x^2}}}} \right)$ and g(x) is nothing but differentiation of y with respect to x.

Complete step by step solution:
Given: Let us assume the function to be y
Therefore,
$y = {\tan ^{ - 1}}\left( {\dfrac{{6x\sqrt x }}{{1 - 9{x^3}}}} \right)$
Which can also be written as

$${\tan ^{ - 1}}\left[ {\dfrac{{2x,3{x^{3/2}}}}{{1 - {{\left( {3{x^{3/2}}} \right)}^2}}}} \right] \\\ y = 2{\tan ^{ - 1}}\left( {3{x^{3/2}}} \right) ......\left( i \right) \\\$$

Since we knew the formula of $\phi \,\,2{\tan ^{ - 1}}x$ were well that is $\phi$ nothing by $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{2 \times {x^2}}}{{1 - {x^2}}}} \right)$ thus we got the assume.
Now, differentiating y with respect to x we get.

$$\dfrac{{dy}}{{dx}} = 2.{\tan ^{ - 1}}\left( {3{x^{3/2}}} \right) \\\ \, = 2.\dfrac{1}{{1 + {{\left( {3{x^{3/2}}} \right)}^2}}}.3 \times \dfrac{3}{2} \times {\left( x \right)^{1/2}} \\\ \dfrac{{dy}}{{dx}} = \dfrac{9}{{1 + 9{x^3}}}.\sqrt x \\\$$

Thus we got the value of $\dfrac{{dy}}{{dx}}$ and according to question this is nothing but $g\left( x \right)$ thus we can state that the value of $g\left( x \right)$ as $\dfrac{9}{{1 + 9{x^3}}}.\sqrt x$

Note: In this type of question students often make mistakes while breaking the variable into any subsequent formula. They break it or form it into a pattern such that it can be easily transformed into standard formula. If a student’s tackle this problem then the rest of the problem is just bread and bester. Also, do not make silly mistakes like performing differentiation as they require very intones calculation.