Question:
If $ y={{\sin }^{-1}}\sqrt{1-x}, $ then $ \frac{dy}{dx} $ is equal to
If $ y={{\sin }^{-1}}\sqrt{1-x}, $ then $ \frac{dy}{dx} $ is equal to
Updated On: Jun 7, 2024
- $ \frac{1}{\sqrt{1-x}} $
- $ \frac{-1}{2\sqrt{1-x}} $
- $ \frac{1}{\sqrt{x}} $
- $ \frac{-1}{2\sqrt{x}\sqrt{1-x}} $
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The Correct Option is D
Solution and Explanation
Given that, $ y={{\sin }^{-1}}\sqrt{1-x} $
Differentiating w.r.t. $ x, $ we have
$ \frac{dy}{dx}=\frac{1}{\sqrt{1-(1-x)}}.\frac{1}{2}.\frac{1}{\sqrt{1-x}}.(-1) $
$=\frac{1}{\sqrt{x}}.\frac{(-1)}{2\sqrt{1-x}} $
$=\frac{(-1)}{2\sqrt{x}.\sqrt{1-x}} $
Differentiating w.r.t. $ x, $ we have
$ \frac{dy}{dx}=\frac{1}{\sqrt{1-(1-x)}}.\frac{1}{2}.\frac{1}{\sqrt{1-x}}.(-1) $
$=\frac{1}{\sqrt{x}}.\frac{(-1)}{2\sqrt{1-x}} $
$=\frac{(-1)}{2\sqrt{x}.\sqrt{1-x}} $
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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.