Question

If $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\dfrac{{4\sqrt x }}{{1 - 4x}}} \right) =$
A.$\dfrac{1}{{\sqrt x \left( {1 + 4x} \right)}}$
B.$\dfrac{2}{{\sqrt x \left( {1 + 4x} \right)}}$
C.$\dfrac{4}{{\sqrt x \left( {1 + 4x} \right)}}$
D.None of these

Answer 1

Hint : Here, the given question. We have to find the derivative or differentiated term of the function. For this, first consider a given trigonometric expression as $y$, then differentiate $y$ with respect to $x$ by using a standard differentiation formula and use quotient rule for differentiation then on further simplification we get the required differentiation value.

Complete step-by-step answer :
Differentiation can be defined as a derivative of a function with respect to an independent variable
Otherwise
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Let $y = f\left( x \right)$ be a function of. Then, the rate of change of “y” per unit change in “x” is given by $\dfrac{{dy}}{{dx}}$.
The quotient rule is used to differentiate many functions where one function is divided by another. The formal definition of the rule is: $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\left( {\dfrac{{du}}{{dx}}} \right) - u\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$.
Consider the given function as $y$
$y = {\tan ^{ - 1}}\dfrac{{4\sqrt x }}{{1 - 4x}}$ ------- (1)
Multiply ‘$\tan$’ on both side
$\tan y = \tan \left( {{{\tan }^{ - 1}}\dfrac{{4\sqrt x }}{{1 - 4x}}} \right)$
On simplification, we get
$\tan y = \dfrac{{4\sqrt x }}{{1 - 4x}}$ -----(2)
Now we have to differentiate this function on both sides with respect to $x$.
$\Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( {\tan y} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{4\sqrt x }}{{1 - 4x}}} \right)$
Apply a quotient rule of differentiation on RHS, then
$\Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( {\tan y} \right) = \dfrac{{\left( {1 - 4x} \right) \cdot \left( {\dfrac{d}{{dx}}4\sqrt x } \right) - 4\sqrt x \cdot \left( {\dfrac{d}{{dx}}\left( {1 - 4x} \right)} \right)}}{{{{\left( {1 - 4x} \right)}^2}}}$ -----(3)
Using the standard differentiated formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, $\dfrac{d}{{dx}}\left( {\sqrt x } \right) = \dfrac{1}{{2\sqrt x }}$ and $\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$, then equation (3) becomes
$\Rightarrow \,\,\,\,\left( {{{\sec }^2}y} \right)\dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 - 4x} \right) \cdot \left( {4\dfrac{1}{{2\sqrt x }}} \right) - 4\sqrt x \cdot \left( { - 4} \right)}}{{{{\left( {1 - 4x} \right)}^2}}}$
We know the trigonometric identity $1 + {\tan ^2}\theta = {\sec ^2}\theta$, then we have
$\Rightarrow \,\,\,\,\left( {1 + {{\tan }^2}y} \right)\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{2}{{\sqrt x }}\left( {1 - 4x} \right) + 16\sqrt x }}{{{{\left( {1 - 4x} \right)}^2}}}$
$\Rightarrow \,\,\,\,\left( {1 + {{\tan }^2}y} \right)\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{2}{{\sqrt x }} - 8\sqrt x + 16\sqrt x }}{{{{\left( {1 - 4x} \right)}^2}}}$
$\Rightarrow \,\,\,\,\left( {1 + {{\tan }^2}y} \right)\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{2}{{\sqrt x }} + 8\sqrt x }}{{{{\left( {1 - 4x} \right)}^2}}}$
$\Rightarrow \,\,\,\,\left( {1 + {{\tan }^2}y} \right)\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{2 + 8{{\left( {\sqrt x } \right)}^2}}}{{\sqrt x }}}}{{{{\left( {1 - 4x} \right)}^2}}}$
$\Rightarrow \,\,\,\,\left( {1 + {{\tan }^2}y} \right)\dfrac{{dy}}{{dx}} = \dfrac{{2 + 8x}}{{\sqrt x {{\left( {1 - 4x} \right)}^2}}}$
Divide both side by $\left( {1 + {{\tan }^2}y} \right)$, then we have
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{2 + 8x}}{{\sqrt x {{\left( {1 - 4x} \right)}^2}}} \times \dfrac{1}{{1 + {{\tan }^2}y}}$
Substitute $\tan y = \dfrac{{4\sqrt x }}{{1 - 4x}}$, then
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{2 + 8x}}{{\sqrt x {{\left( {1 - 4x} \right)}^2}}} \times \dfrac{1}{{1 + {{\left( {\dfrac{{4\sqrt x }}{{1 - 4x}}} \right)}^2}}}$
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{2 + 8x}}{{\sqrt x {{\left( {1 - 4x} \right)}^2}}} \times \dfrac{1}{{\dfrac{{{{\left( {1 - 4x} \right)}^2} + {{\left( {4\sqrt x } \right)}^2}}}{{{{\left( {1 - 4x} \right)}^2}}}}}$
On simplification, we get
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{2 + 8x}}{{\sqrt x \left( {{{\left( {1 - 4x} \right)}^2} + {{\left( {4\sqrt x } \right)}^2}} \right)}}$
Apply a identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{2 + 8x}}{{\sqrt x \left( {1 + 16{x^2} - 8x + 16x} \right)}}$
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{2 + 8x}}{{\sqrt x \left( {1 + 16{x^2} + 8x} \right)}}$
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{2 + 8x}}{{\sqrt x {{\left( {1 + 4x} \right)}^2}}}$
Take 2 as common in the numerator of RHS
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{2\left( {1 + 4x} \right)}}{{\sqrt x {{\left( {1 + 4x} \right)}^2}}}$
On cancelling a like terms $\left( {1 + 4x} \right)$ in both numerator and denominator, then we get
$\therefore \,\,\,\,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{2}{{\sqrt x \left( {1 + 4x} \right)}}$
Hence, the required the differentiated value $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\dfrac{{4\sqrt x }}{{1 - 4x}}} \right) = \dfrac{2}{{\sqrt x \left( {1 + 4x} \right)}}$.
Therefore, option (B) is the correct answer.
So, the correct answer is “Option B”.

Note : When differentiating the function or term the student must recognize the dependent and independent variable then differentiate the dependent variable with respect to the independent variable. Should know the product and quotient rule of differentiation and remember the standard differentiation formulas.