Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Differentiate of the Following from First Principle:Sin (X + 1) - Mathematics

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Differentiate  of the following from first principle:

sin (x + 1)

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Solution

\[ \frac{d}{dx}\left( f\left( x \right) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[\frac{d}{dx}\left( \sin \left( x + 1 \right) \right) = \lim_{h \to 0} \frac{\sin \left( x + h + 1 \right) - \sin \left( x + 1 \right)}{h}\]
\[\text{ We know }:\]
\[\sin C - \sin D = 2 \cos \left( \frac{C + D}{2} \right) \sin \left( \frac{C - D}{2} \right)\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{x + h + 1 + x + 1}{2} \right) \sin \left( \frac{x + h + 1 - x - 1}{2} \right)}{h}\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{2x + h + 2}{2} \right) \sin \left( \frac{h}{2} \right)}{h}\]
\[ = 2 \lim_{h \to 0} \cos \left( \frac{2x + h + 2}{2} \right) \lim_{h \to 0} \frac{\sin \left( \frac{h}{2} \right)}{\frac{h}{2}} \times \frac{1}{2}\]
\[ = 2 \cos \left( x + 1 \right) \times \frac{1}{2}\]
\[ = \cos \left( x + 1 \right)\]

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 30.2 | Q 2.07 | Page 25

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