Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
Advertisement Remove all ads

Differentiate of the Following from First Principle: Sin (2x − 3) - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads

Differentiate  of the following from first principle:

sin (2x − 3)

Advertisement Remove all ads

Solution

\[\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\sin \left( 2x + 2h - 3 \right) - \sin \left( 2x - 3 \right)}{h}\]
\[\text{ We know }:\]
\[sin C-sin D=2 cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right)\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{2x + 2h - 3 + 2x - 3}{2} \right) \sin \left( \frac{2x + 2h - 3 + 2x - 3}{2} \right)}{h}\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{4x + 2h - 6}{2} \right) \sin \left( h \right)}{h}\]
\[ = \lim_{h \to 0} 2 \cos \left( \frac{4x + 2h - 6}{2} \right) \lim_{h \to 0} \frac{\sin h}{h}\]
\[ = 2 \cos \left( \frac{4x - 6}{2} \right) \left( 1 \right)\]
\[ = 2 \cos \left( 2x - 3 \right)\]
\[ \]
\[\]

Concept: The Concept of Derivative - Algebra of Derivative of Functions
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.2 | Q 2.11 | Page 25

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×