Question
Differentiate the following functions with respect to x:
$\text{e}^\text{x}\log\sin2\text{x}$
$\text{e}^\text{x}\log\sin2\text{x}$
✓
Answer
Let $\text{y}=\text{e}^\text{x}\log\sin2\text{x}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{e}^\text{x}\log\sin2\text{x}\big]$
$=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}\log\sin2\text{x}+\log\sin2\text{x}\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\big)$
[Using product rule and chain rule]
$=\text{e}^\text{x}\frac{1}{\sin2\text{x}}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\log\sin2\text{x}\big(\text{e}^\text{x}\big)$
$=\frac{\text{e}^\text{x}}{\sin2\text{x}}\cos2\text{x}\frac{\text{d}}{\text{dx}}(2\text{x})+\text{e}^\text{x}\log\sin2\text{x}$
$=\frac{2\cos2\text{xe}^\text{x}}{\sin2\text{x}}+\text{e}^\text{x}\log\sin2\text{x}$
$=\text{e}^\text{x}(2\cot2\text{x}+\log\sin2\text{x})$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\log\sin2\text{x}\big)=\text{e}^\text{x}(2\cot2\text{x}+\log\sin2\text{x})$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{e}^\text{x}\log\sin2\text{x}\big]$
$=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}\log\sin2\text{x}+\log\sin2\text{x}\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\big)$
[Using product rule and chain rule]
$=\text{e}^\text{x}\frac{1}{\sin2\text{x}}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\log\sin2\text{x}\big(\text{e}^\text{x}\big)$
$=\frac{\text{e}^\text{x}}{\sin2\text{x}}\cos2\text{x}\frac{\text{d}}{\text{dx}}(2\text{x})+\text{e}^\text{x}\log\sin2\text{x}$
$=\frac{2\cos2\text{xe}^\text{x}}{\sin2\text{x}}+\text{e}^\text{x}\log\sin2\text{x}$
$=\text{e}^\text{x}(2\cot2\text{x}+\log\sin2\text{x})$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\log\sin2\text{x}\big)=\text{e}^\text{x}(2\cot2\text{x}+\log\sin2\text{x})$
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