The derivative of ${\text{co}}{{\text{s}}^{ - 1}}\left( {{\text{2}}{{\text{x}}^2} - 1} \right)$with respect to ${\text{co}}{{\text{s}}^{ - 1}}{\text{x}}$is: ${\text{A}}{\text{. }}\dfrac{{ - 1}}{{2\sqrt {1 - {{\text{x}}^2}} }} \\ {\text{B}}{\text{. }}\dfrac{2}{{\text{x}}} \\ {\text{C}}{\text{. 2}} \\ {\text{D}}{\text{. 1 - }}{{\text{x}}^2} \\$

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Hint: In order to find the derivative of one inverse trigonometric function w.r.t another, we perform the differentiation in two parts. We differentiate the first and second terms w.r.t ‘x’ individually and then divide the answers obtained with each other.

Given Data,
${\text{co}}{{\text{s}}^{ - 1}}\left( {{\text{2}}{{\text{x}}^2} - 1} \right)$
${\text{co}}{{\text{s}}^{ - 1}}{\text{x}}$
Let us consider the functions, ${\text{co}}{{\text{s}}^{ - 1}}\left( {{\text{2}}{{\text{x}}^2} - 1} \right)$= y and ${\text{co}}{{\text{s}}^{ - 1}}{\text{x}}$= z
According to the question we are supposed to find, $\dfrac{{{\text{dy}}}}{{{\text{dz}}}}$
We find this as follows, we differentiate y and z w.r.t ‘x’ individually and then divide them,
$\dfrac{{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}}}{{\dfrac{{{\text{dz}}}}{{{\text{dx}}}}}} = \dfrac{{{\text{dy}}}}{{{\text{dz}}}}$ .

Now $\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{co}}{{\text{s}}^{ - 1}}\left( {{\text{2}}{{\text{x}}^2} - 1} \right)} \right)$
$\Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{ - 1}}{{\sqrt {1 - {{\left( {2{{\text{x}}^2} - 1} \right)}^2}} }} \times \left[ {{\text{4x - 0}}} \right]$
$\Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{ - 1 \times {\text{4x}}}}{{\sqrt {1 - 4{{\text{x}}^4} - 1 + 4{{\text{x}}^2}} }}$
$\Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{ - {\text{4x}}}}{{{\text{2x}}\sqrt {1 - {{\text{x}}^2}} }}$
$\Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{ - 2}}{{\sqrt {1 - {{\text{x}}^2}} }}{\text{ - - - }}\left( 1 \right)$

Now $\dfrac{{{\text{dz}}}}{{{\text{dx}}}} = \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{co}}{{\text{s}}^{ - 1}}\left( {\text{x}} \right)} \right)$
$\Rightarrow \dfrac{{{\text{dz}}}}{{{\text{dx}}}} = \dfrac{{ - 1}}{{\sqrt {1 - {{\text{x}}^2}} }}{\text{ - - - }}\left( 2 \right)$

From equation (1) and (2) we can write,
$\Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dz}}}} = \dfrac{{ - 2/\sqrt {1 - {{\text{x}}^2}} }}{{ - 1/\sqrt {1 - {{\text{x}}^2}} }} = 2$
So, the correct answer is “Option C”.

Note: In order to solve this type of problem the key is to know how to differentiate two terms when both the terms are complicated. This method can be used to simplify terms of any order. The value of differentiation of an inverse trigonometric function can be found by looking at the differentiations of common trigonometric terms. Good knowledge in performing differentiation operations on general terms and the respective formulae is needed to solve these problems.