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Question:
If $ y={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{x}^{2}}}} \right), $ then $ \frac{dy}{dx} $ is equal to
If $ y={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{x}^{2}}}} \right), $ then $ \frac{dy}{dx} $ is equal to
Updated On: Jun 23, 2024
- $ \frac{1}{\sqrt{1-{{x}^{2}}}} $
- $ \frac{2}{\sqrt{1-{{x}^{2}}}} $
- $ \frac{1}{\sqrt{1+{{x}^{2}}}} $
- $ \frac{2}{\sqrt{1+{{x}^{2}}}} $
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The Correct Option is A
Solution and Explanation
Given, $ y={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{x}^{2}}}} \right) $
Put, $ x=\sin \theta $
$ \therefore $ $ y={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{\sin }^{2}}\theta }} \right) $
$ \Rightarrow $ $ y={{\sec }^{-1}}(\sec \theta )=\theta $
$ \Rightarrow $ $ y={{\sin }^{-1}}x $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}} $
Put, $ x=\sin \theta $
$ \therefore $ $ y={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{\sin }^{2}}\theta }} \right) $
$ \Rightarrow $ $ y={{\sec }^{-1}}(\sec \theta )=\theta $
$ \Rightarrow $ $ y={{\sin }^{-1}}x $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}} $
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.
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