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A + Sin X 1 + a Sin X - Mathematics

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\[\frac{a + \sin x}{1 + a \sin x}\] 

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Solution

Let us use the quotient rule here.
We have:
u = a + sin x and v =1 + a sin x
u' = cos x and v'=a cos 

\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{a + \sin x}{1 + a\sin x} \right) = \frac{(1 + a\sin x)(\cos x) - (a + \sin x)(a\cos x)}{(1 + a\sin x )^2}\]
\[ = \frac{\cos x + a\sin x \cos x - a^2 \cos x - a \sin x \cos x}{(1 + a\sin x )^2}\]
\[ = \frac{\cos x - a^2 \cos x}{(1 + a\sin x )^2}\]
\[ = \frac{(1 - a^2 )\cos x}{(1 + a \sin x )^2}\] 

 

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.5 | Q 16 | Page 44

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