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**Hint:**For solving this particular question we must know that Differentiation of trigonometric function sine of variable $x$ is equal to cosine of variable $x$ , and differentiation of $\sin ax = a\cos ax$ where $a$ is any constant and $x$ is the variable .

**Complete solution step by step:**

It is given it the question that ,

\[y = \sin 2x\] (given)

Now , differentiate on both sides with respect to $x$ , we will get ,

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(\sin 2x)$

$ \Rightarrow \dfrac{{dy}}{{dx}} = \cos (2x).2$ (since differentiation of $\sin ax = a\cos ax$ )

$ = 2\cos 2x$

Therefore , $\dfrac{{dy}}{{dx}} = 2\cos 2x$ .

**And we can say that option A is the correct option.**

**Formula used:**For solving this particular question we used ,

$\dfrac{d}{{dx}}(\sin ax) = a\cos ax$ ,where $a$is any constant and $x$ is the variable .

Differentiation of trigonometric function sine of variable $x$ is equal to cosine of variable $x$ .

**Additional Information:**

Differentiation is the essence of Calculus. A derivative is defined because the instantaneous rate of change in function supported one in every of its variables. it's just like finding the slope of a tangent to the function at some extent. we've got following rules in differentiation ,

• Sum and Difference rule that claims (u(x) ± v(x))’=u'(x)±v'(x)

• Product rule that says (u(x) × v(x))’=u′(x)×v(x)+u(x)×v′(x)

• Quotient Rule that claims (u(x)/v(x))’ =(u′(x)×v(x)−u(x)×v′(x))/(v(x))2

• Chain Rule that says dy(u(x))/dx = dy/du × du/dx

**Note:**The differentiation of a function $f(x)$ is represented as $f'(x)$ . If $f(x) = y$, then $f'(x) =\dfrac{{dy}}{{dx}}$ , which means $y$ is differentiated with respect to $x$ Differentiation of trigonometric function sine of variable $x$ is equal to cosine of variable $x$ , and differentiation of $\sin ax = a\cos ax$ where $a$ is any constant and $x$ is the variable

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