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Question: If\(y = {\sec ^{ - 1}}\left( {\dfrac{1}{{1 - 2{x^2}}}} \right)\), then\(\dfrac{{dy}}{{dx}}\) A. \(...

Ify=sec⁡−1(11−2x2)y = {\sec ^{ - 1}}\left( {\dfrac{1}{{1 - 2{x^2}}}} \right), thendydx\dfrac{{dy}}{{dx}}
A. 11−x2\dfrac{1}{{\sqrt {1 - {x^2}} }}
B. 21−x2\dfrac{2}{{\sqrt {1 - {x^2}} }}
C. 11+x2\dfrac{1}{{\sqrt {1 + {x^2}} }}
D. 21+x2\dfrac{2}{{\sqrt {1 + {x^2}} }}

Explanation

Solution

First, we shall analyze the given information so that we are able to solve the problem. In Mathematics, the derivative refers to the rate of change of a function with respect to a variable. Here in this question, we are asked to calculate the first derivative of the given equation. First, we need to change the given equation for our convenience. To change the equation, we need to apply some suitable trigonometric identities. Then, we need to differentiate the resultant equation.
Formula to be used:
a) The trigonometric identities that we need to apply in this problem are as follows.
1−2sin⁡2θ=cos⁡2θ1 - 2{\sin ^2}\theta = \cos 2\theta
1cos⁡θ=sec⁡θ\dfrac{1}{{\cos \theta }} = \sec \theta
sec⁡−1sec⁡x=x{\sec ^{ - 1}}\sec x = x
b) The derivative formula that we need to apply in this problem is as follows.
ddx(sin⁡−1x)=11−x2\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}

Complete step by step answer:
It is given thaty=sec⁡−1(11−2x2)y = {\sec ^{ - 1}}\left( {\dfrac{1}{{1 - 2{x^2}}}} \right) .
To find:dydx\dfrac{{dy}}{{dx}}
Now, let's put x=sin⁡θx = \sin \theta it in the given equation.
⇒θ=sin⁡−1x\Rightarrow \theta = {\sin ^{ - 1}}x …..(1)\left( 1 \right)
Thus,y=sec⁡−1(11−2x2)y = {\sec ^{ - 1}}\left( {\dfrac{1}{{1 - 2{x^2}}}} \right)
=sec⁡−1(11−2sin⁡2θ)= {\sec ^{ - 1}}\left( {\dfrac{1}{{1 - 2{{\sin }^2}\theta }}} \right)
=sec⁡−1(1cos⁡2θ)= {\sec ^{ - 1}}\left( {\dfrac{1}{{\cos 2\theta }}} \right) (Here we applied the trigonometric identity 1−2sin⁡2θ=cos⁡2θ1 - 2{\sin ^2}\theta = \cos 2\theta )
=sec⁡−1sec⁡2θ= {\sec ^{ - 1}}\sec 2\theta (Here we applied the trigonometric identity1cos⁡θ=sec⁡θ\dfrac{1}{{\cos \theta }} = \sec \theta )
=2θ= 2\theta (Here we appliedsec⁡−1sec⁡x=x{\sec ^{ - 1}}\sec x = x )
=2sin⁡−1x2{\sin ^{ - 1}}x (Here we applied the equation (1)\left( 1 \right)
Hence,y=2sin⁡−1xy = 2{\sin ^{ - 1}}x
Now, we shall differentiate the above equation with respect to xx .
dydx=ddx(2sin⁡−1x)\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(2{\sin ^{ - 1}}x)
=2ddx(sin⁡−1x)= 2\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right)
=21−x2= \dfrac{2}{{\sqrt {1 - {x^2}} }} (Here we appliedddx(sin⁡−1x)=11−x2\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }} )
Thus, we get dydx=21−x2\dfrac{{dy}}{{dx}} = \dfrac{2}{{\sqrt {1 - {x^2}} }}

So, the correct answer is “Option B”.

Note: When we are asked to find the derivation of the given equation, we need to change the given equation smartly for our convenience. Here we have applied trigonometric identities to change the equation. Then we need to analyze where we need to apply the derivative formulae and where we need to apply the rule of differentiation while differentiating the given equation.