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explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 43 maths

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Answer:  \sin \{2 \log (2 x+3)\}\left(\frac{2}{(2 x+3)}\right)

Hint: you must know the rules of solving derivative of trigonometric and logarithm function,

Given: \sin ^{2}\{\log (2 x+3)\}

Solution:

Let  y=\sin ^{2}\{\log (2 x+3)\}

Differentiate with respect to x,

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left[\sin ^{2} \log (2 x+3)\right] \\ \\&=2 \sin \{\log (2 x+3)\} \frac{d}{d x} \sin \{\log (2 x+3)\} \end{aligned}

\begin{aligned} &=2 \sin \{\log (2 x+3)\} \cos \{\log (2 x+3)\} \frac{\mathrm{d}}{\mathrm{dx}} \log (2 x+3) \\\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\sin \{2 \log (2 x+3)\}\left(\frac{2}{(2 x+3)}\right) \end{aligned}

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