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Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 23

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Answer: \frac{d y}{d x}=e^{3 x} \cdot \sin 4 x .2 x\left[e^{3 x} \cdot \sin 4 x \cdot 2^{x}(3+4 \cot 4 x+\ln 2)\right]

Hint: Differentiate the equation taking log on both sides

Given: y=e^{3 x} \cdot \sin 4 x \cdot 2^{x}


                                            [\because \log A B=\log A+\log B]

        \ln y=\ln e^{3 x}+\ln \sin 4 x+\ln 2^{x}

Diff w.r.t x

        \frac{1}{y} \frac{d y}{d x}=\frac{1}{e^{3 x}} \cdot \cdot e^{3 x} \cdot 3+\frac{\cos 4 x}{\sin 4 x} \cdot 4+\ln 2

        \frac{d y}{d x}=e^{3 x} \cdot \sin 4 x .2 x\left[e^{3 x} \cdot \sin 4 x .2 x(3+4 \cot 4 x+\ln 2)\right]

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