Find dy dx y=e^ 3x 4x 2^x - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{e}^{3\text{x}}\sin4\text{x}\times2^\text{x}$

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Answer

We have, $\text{y}=\text{e}^{3\text{x}}\sin4\text{x}\times2^\text{x}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log\text{e}^{3\text{x}}+\log\sin4\text{x}+\log2^\text{x}$
$\Rightarrow\log\text{y}=3\text{x}\log\text{e}+\log\sin4\text{x}+\text{x}\log2$
$\Rightarrow\log\text{y}=3\text{x}+\log\sin4\text{x}+\text{x}\log2$
Differentiating with resepect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{\text{d}}{\text{dx}}(\sin4\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}\log2)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\frac{1}{\sin4\text{x}}\frac{\text{d}}{\text{dx}}(\sin4\text{x})+\log2(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\frac{1}{\sin4\text{x}}(\cos4\text{x})\frac{\text{d}}{\text{dx}}(4\text{x})+\log2$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+\cot4\text{x}(4)+\log2$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=3+4\cot4\text{x}+\log2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[3+4\cot4\text{x}+\log2\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{3\text{x}}\sin4\text{x}2^\text{x}\big[3+4\cot4\text{x}+\log2\big]$
[Using equation (i)]

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