Question

What is the derivative of \[\sin 2(x)\] by first principle?

Answer 1

Hint:
> The right hand derivative of \[f\left( x \right)\] at \[x = a\] denoted by \[f'({a^ + })\] is defined as:
\[f'({a^ + }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}\], provided the limit exists and is finite.
> The left hand derivative of \[f\left( x \right)\] at \[x = a\] denoted by \[f'({a^ - })\] is defined as:
\[f'({a^ - }) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a - h) - f(a)}}{{ - h}}\], provided the limit exists and is finite.
The first principle of differentiation is the process of using the above formula for calculating the derivative. The first principle of differentiation is also known as Delta method.
Given: \[f(x) = \sin 2(x)\]
To find: The derivative of \[\sin 2(x)\] by first principle

Complete step-by-step solution:
Step 1: As we know, according to the first principle if f is a real function and c is a point in its domain. The derivative of f at c is defined by
\[\mathop {\lim }\limits_{h \to 0} \dfrac{{f(c + h) - f(c)}}{h}\] Or \[\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\] provided this limit exist.
Substituting the value of \[f(x)\] and \[f(x + h)\] we get
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin 2(x + h) - \sin 2(x)}}{h}\]
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (2x + 2h) - \sin 2(x)}}{h}\]
Step 2: from trigonometric properties we know that
\[\sin C - \sin D = 2\cos \dfrac{{C + D}}{2}\sin \dfrac{{C - D}}{2}\]
Using above trigonometric property, we can write
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (2x + 2h) - \sin 2(x)}}{h}\]
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2\cos (2x + h)\sinh }}{h}\]
Step 3: from limit of trigonometric function we know that \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\]
Therefore,
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2\cos (2x + h)\sinh }}{h}\]
\[f'(x) = \mathop {\lim }\limits_{h \to 0} 2\cos (2x + h).\mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h}\]
\[f'(x) = \mathop {\lim }\limits_{h \to 0} 2\cos (2x + h)\]
Applying we have h tends to zero, we get
\[f'(x) = 2\cos (2x)\]
Hence, the derivative of \[\sin 2(x)\] by first principle is equal to \[2\cos (2x)\].

Note:
> In \[\mathop {\lim }\limits_{x \to a} f(x),x \to a\] necessarily implies \[x \ne a\]. That is while evaluating limits at x=a, we are not concerned with the value of the function at x=a. in fact the function may or may not be defined at x=a.
> If \[f(x)\] is defined only on one side of \[x = a\], one sided limits are good enough to establish the existence of limits, & if \[f(x)\] is defined on either side of ‘a’ both sided limits are to be considered