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Question: The derivative of \[{\text{co}}{{\text{s}}^{ - 1}}\left( {{\text{2}}{{\text{x}}^2} - 1} \right)\]wit...

The derivative of cos−1(2x2−1){\text{co}}{{\text{s}}^{ - 1}}\left( {{\text{2}}{{\text{x}}^2} - 1} \right)with respect to cos−1x{\text{co}}{{\text{s}}^{ - 1}}{\text{x}}is:
A. −121−x2 B. 2x C. 2 D. 1 - x2  {\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} \\\

Explanation

Solution

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.

Complete step-by-step answer :
Given Data,
cos−1(2x2−1){\text{co}}{{\text{s}}^{ - 1}}\left( {{\text{2}}{{\text{x}}^2} - 1} \right)
cos−1x{\text{co}}{{\text{s}}^{ - 1}}{\text{x}}
Let us consider the functions, cos−1(2x2−1){\text{co}}{{\text{s}}^{ - 1}}\left( {{\text{2}}{{\text{x}}^2} - 1} \right)= y and cos−1x{\text{co}}{{\text{s}}^{ - 1}}{\text{x}}= z
According to the question we are supposed to find, dydz\dfrac{{{\text{dy}}}}{{{\text{dz}}}}
We find this as follows, we differentiate y and z w.r.t ‘x’ individually and then divide them,
dydxdzdx=dydz\dfrac{{\dfrac{{{\text{dy}}}}{{{\text{dx}}}}}}{{\dfrac{{{\text{dz}}}}{{{\text{dx}}}}}} = \dfrac{{{\text{dy}}}}{{{\text{dz}}}} .

Now dydx=ddx(cos−1(2x2−1))\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)
⇒dydx=−11−(2x2−1)2×[4x - 0]\Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{ - 1}}{{\sqrt {1 - {{\left( {2{{\text{x}}^2} - 1} \right)}^2}} }} \times \left[ {{\text{4x - 0}}} \right]
⇒dydx=−1×4x1−4x4−1+4x2\Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{ - 1 \times {\text{4x}}}}{{\sqrt {1 - 4{{\text{x}}^4} - 1 + 4{{\text{x}}^2}} }}
⇒dydx=−4x2x1−x2\Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{ - {\text{4x}}}}{{{\text{2x}}\sqrt {1 - {{\text{x}}^2}} }}
⇒dydx=−21−x2 - - - (1)\Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{ - 2}}{{\sqrt {1 - {{\text{x}}^2}} }}{\text{ - - - }}\left( 1 \right)

Now dzdx=ddx(cos−1(x))\dfrac{{{\text{dz}}}}{{{\text{dx}}}} = \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{co}}{{\text{s}}^{ - 1}}\left( {\text{x}} \right)} \right)
⇒dzdx=−11−x2 - - - (2)\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,
⇒dydz=−2/1−x2−1/1−x2=2\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.