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  • Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 24 maths

Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 24 maths

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Answer:  \frac{d y}{d x}=\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \sin 4 x[\cot x+\cot 2 x .2+3 \cot 3 x+4 \cot 4 x]

Hint: Differentiate the equation taking log on both sides

Given: y=\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \sin 4 x

Solution:  

        y=\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \sin 4 x

Taking log on both sides,

        \log y=\log (\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \sin 4 x)

                  =\log \sin x+\log \sin 2 x+\log \sin 3 x+\log \sin 4 x

        \frac{1}{y} \frac{d y}{d x}=\cot x+\cot 2 x .2+3 \cot 3 x+4 \cot 4 x\left[\because \frac{d}{d x}(\log \sin x)=\frac{1}{\sin x} \cdot \cos x=\cot x\right]

        \frac{d y}{d x}=y[\cot x+\cot 2 x .2+3 \cot 3 x+4 \cot 4 x]

        \frac{d y}{d x}=\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \sin 4 x[\cot x+\cot 2 x .2+3 \cot 3 x+4 \cot 4 x]

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