Find the derivative of sin 2x by first principle.

1.	Find the derivative of sin 2x by first principle.
Answer» Let f(x) = sin 2x ∴ By first principle \(f'(x)= \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\) \(= \lim\limits_{h \to 0}\frac{sin2(x+h)-sin2x}{h}\) \(= \lim\limits_{h \to 0}\frac{2cos\frac{2(x+h)+2x}{2}sin\frac{2(x+h)-2x}{2}}{h}\) = \( \lim\limits_{h \to 0}\frac{2cos(h+2x)sinh}{h}\) = 2\(\lim\limits_{h \to0 }cos(h+2x)\lim\limits_{h \to 0}(\frac{sinh}{h})\) = 2∙ cos 2x ∙1 = 2 cos 2x..

Answer»

Let f(x) = sin 2x

∴ By first principle

\(f'(x)= \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\)

\(= \lim\limits_{h \to 0}\frac{sin2(x+h)-sin2x}{h}\)

\(= \lim\limits_{h \to 0}\frac{2cos\frac{2(x+h)+2x}{2}sin\frac{2(x+h)-2x}{2}}{h}\)

= \( \lim\limits_{h \to 0}\frac{2cos(h+2x)sinh}{h}\)

= 2\(\lim\limits_{h \to0 }cos(h+2x)\lim\limits_{h \to 0}(\frac{sinh}{h})\)

= 2∙ cos 2x ∙1

= 2 cos 2x..

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