Question:
Differentiate $ \sin \,\,(\sin \,\,2x). $
Differentiate $ \sin \,\,(\sin \,\,2x). $
Updated On: Jun 23, 2024
- $ 2\,\cos \,2x.\,\cos \,2x $
- $ 2\,\cos \,2x.\,\cos \,(\sin \,2x) $
- $ 2\,\cos \,2x.\sin 2x $
- $ \cos \,2x.\cos \,(\sin 2x) $
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The Correct Option is B
Solution and Explanation
Let $ y=\sin (\sin \,2x) $ Then, $ \frac{dy}{dx}=\frac{d}{dx}\,\,[\sin \,(\sin 2x)] $
$=\cos (\sin 2x).\cos \,2x.2 $ $ \frac{dy}{dx}=2\cos \,2x.\cos (\sin 2x) $
$=\cos (\sin 2x).\cos \,2x.2 $ $ \frac{dy}{dx}=2\cos \,2x.\cos (\sin 2x) $
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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.