Question

If $y=\sin \left( 2x \right)$ , then $\dfrac{dy}{dx}$ is equal to
(a) $2\cos \left( 2x \right)$
(b) $2\cos \left( x \right)$
(c) $2\sin \left( x \right)$
(d) $2\sin \left( 2x \right)$

Answer 1

Hint: First expand the given expression by using general trigonometric identities. Then apply the u-v rule in the differentiative u – v rule implies (u.v) rule.
\[\dfrac{d}{dx}\left( f\circ g \right)=\left( {f}'\circ g \right).\left( {{g}'} \right)\]

Complete step-by-step answer:
Given Expression from the question:
$y=\sin \left( 2x \right)$
Chain rule of differentiation: In calculus the chain rule is a formula to compute the derivative of a composite function. That is the composite of f and g is fog. If f and g are differentiable functions then the chain rule tells that derivative of their composite function is given by:
\[\dfrac{d}{dx}\left( f\circ g \right)=\left( {f}'\circ g \right).\left( {{g}'} \right)\]
By using chain rule of differentiation described above, we have values of f and g as:
Here we know (f o g) (x) = sin(2x)
So, we get :
f (x) = sin (x).
g (x) = 2x.
by basic knowledge of algebra, we can say that:
sin (x) and 2x are differentiable functions.
From the above condition we can apply chain rule.
\[{f}'\left( x \right)=\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\sin x\]
By basic knowledge of differentiation we can say value of above expression to be:
d (sinx) = cos(x)
As we know value of g (x), we can say value of \[{f}'\circ g={f}'\left( g\left( x \right) \right)\] to be:
\[\left( {f}'\circ g \right)\left( x \right)=\cos \left( 2x \right)\]
From basic knowledge of differentiation, we can say differentiation of g (x):
\[{g}'\left( x \right)=\dfrac{d}{dx}2x=2\]
From chain rule of differentiation, we can say:
\[\dfrac{d}{dx}\left( f\circ g \right)=\left( {f}'\circ g \right).\left( {{g}'} \right)\]
By substituting values of both terms as calculated above, we get:
\[\dfrac{d}{dx}\left( \sin \left( 2x \right) \right)=\left( \cos \left( 2x \right) \right).\left( 2 \right)\]
By simplifying, we can say value to be 2 cos (2x)
Therefore $2\cos 2x$ is differentiation of given expression.
Option (a) is correct.

Note: Alternative method – 1:
 Directly substitute $2x$ as t in expression and solve with basic trigonometric properties and basic differentiation properties. In that way you will lead to an answer easily and quickly.
Another method – 2:
It is somewhat lengthy process it is done by substituting sin (2x) as 2 sin (x) cos (x) and then apply u.v rule of differentiation:
d(u.v) = u.(dv) + v.(du)