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Question: If \(y={{\sec }^{-1}}\dfrac{2x}{1+{{x}^{2}}}+{{\sin }^{-1}}\dfrac{x-1}{x+1}\) then \(\dfrac{dy}{dx}\...

If y=sec⁡−12x1+x2+sin⁡−1x−1x+1y={{\sec }^{-1}}\dfrac{2x}{1+{{x}^{2}}}+{{\sin }^{-1}}\dfrac{x-1}{x+1} then dydx\dfrac{dy}{dx}=
A. 11
B. x−1x+1\dfrac{x-1}{x+1}
C. Does not exist
D. None of these

Explanation

Solution

This is a question of continuity and differentiation that’s the concept we will be applying here. To find the differentiate of yy we first check if it exists that is check if the equation is continuous and then solve it.

Complete step by step answer:
To solve this question we will start by trying to simplify the current equation so that it is easier to solve the trigonometric function and also get its differentiation which is what we need here. Therefore we know that if we put xx as tan⁡(θ)\tan \left( \theta \right). The value in the first bracket now after putting xx as tan⁡(θ)\tan \left( \theta \right) as we can see is one of the ways we can also usually write sin⁡(2θ)\sin (2\theta ).To show this
Putting x=tan⁡(θ)x=\tan \left( \theta \right), we get
2x1+x2=2tan⁡θ1+tan⁡2θ=sin⁡2θ\dfrac{2x}{1+{{x}^{2}}}=\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta
Now since we know that even though the domain of sin⁡θ\sin \theta is positive infinity to negative infinity the range of sin⁡θ\sin \theta ranges from [−1,1]\left[ -1,1 \right] . Now we must know that for differentiation for any equation to exist the trigonometric function should exist that can be defined at all points for the domain. It can not be that it exists at some points and then fails to exist at others. If that is the case we can say that the differentiation doesn’t exist because the functions that are discontinues in other words are not defined at all points. To check that if the function is continuous we check the range of 2x1+x2\dfrac{2x}{1+{{x}^{2}}} .
Since, −1≤sin⁡θ≤1-1\le \sin \theta \le 1
We can also say that ;
−1≤2x1+x2≤1-1\le \dfrac{2x}{1+{{x}^{2}}}\le 1
Now therefore we can see that first part of this question that is sec⁡−12x1+x2{{\sec }^{-1}}\dfrac{2x}{1+{{x}^{2}}} is only defined at [−1,1]\left[ -1,1 \right] . As stated before we can say that this means that dydx\dfrac{dy}{dx} does not exist.

So, the correct answer is “Option C”.

Note: To explain what a continuous equation is in simple words a function is only continuous when its value exists at all points in the domain. To check that if the derivative exists, we must check the range of the functions, limit, continuity and differentiation of the function.