Question

Differentiate sin² (2x + 1) ?

Answer 1

\[\text{ Let } y = \sin^2 \left( 2x + 1 \right)\]

\[\text{Differentiate it with respect to x we get }, \]

\[\frac{d y}{d x} = \frac{d}{dx}\left[ \sin^2 \left( 2x + 1 \right) \right]\]

\[ = 2\sin\left( 2x + 1 \right)\frac{d}{dx}\sin\left( 2x + 1 \right) \left[ \text{ using chain rule } \right]\]

\[ = 2\sin\left( 2x + 1 \right) \cos\left( 2x + 1 \right) \frac{d}{dx}\left( 2x + 1 \right) \left[ \text{ using chain rule } \right] \]

\[ = 4\sin\left( 2x + 1 \right) \cos\left( 2x + 1 \right)\]

\[ = 2\sin2\left( 2x + 1 \right) \left[ \because \sin2A = 2\sin A\cos A \right]\]

\[ = 2 \sin\left( 4x + 2 \right)\]

\[So, \frac{d}{dx}\left\{ \sin^2 \left( 2x + 1 \right) \right\} = 2 \sin\left( 4x + 2 \right)\]