Question

Differentiate: 2 sin² 2x​

Answer 1

Differentiate :

2 sin² (2x)

Solution :

$\sf \dfrac{d}{dx} \bigg(2 \: \sin ^{2} (2x) \: \bigg) \\ \\ \sf \implies \: 2 \: \bigg( \: \dfrac{d}{dx} \sin ^{2} (2x) \: \bigg) \: \\ \\ \sf \implies \: 2 \: \: \dfrac{d}{dx} \: \bigg( \sin (2x) \: \bigg)^{2}$

[ Note - using identity

$\sf \dfrac{d \: {(x + a)}^{m} }{dx} \: = \: m \times {(x + a)}^{(m - 1)} \times \dfrac{d \: (x + a) }{dx}$ ]

$\sf \implies \: 2 \times \bigg[ 2 \times \: \bigg( \sin (2x) \: ^{(2 - 1)} \bigg) \times \: \dfrac{d( \sin(2x) }{dx} \bigg]\:$

[ Note - Differentiation of sin(x) = cosx ]

$\sf \implies \: 4 \times \: \bigg( \sin (2x) \bigg) \times \bigg( \: \cos(2x) \times \: \dfrac{d(2x) }{dx} \bigg) \\ \\ \sf \implies \:4 \times \sin(2x) \cos(2x) \times \bigg(2 \dfrac{d(x)}{dx} \bigg) \\ \\ \sf \implies \:4 \times \sin(2x) \cos(2x) \times 2 \\ \\ \sf \implies \:8\times \sin(2x) \cos(2x)$

On simplifying,

8 sin(2x)cos(2x) = 2× [ 4 sin(2x) cos(2x) ]

[ Note - use identity , sin(2A) = 2sin(A) cos(A) ]

= 2 × sin(4x)

ANSWER :

$\sf \: \dfrac{d}{dx} \bigg(2 \: \sin^{2} (2x) \: \bigg) \: = \: \bold{8 \sin(2x) \cos(2x) \: = \: \: 2 \sin(4x) }$