Gujarat BoardEnglish MediumSTD 11 ScienceMATHSDerivatives3 Marks
Question
Differentiate the following function with respect to x:$\text{e}^\text{x}\log\sqrt{\text{x}}\tan\text{x}$
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Answer
Let $\text{u}=\text{e}^\text{x};\text{v}=\log\sqrt{\text{x}};\text{w}=\tan\text{x}$Then, $\text{u}'=\text{e}^\text{x};\text{v}'=\frac{1}{\sqrt{\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}=\frac{1}{\text{2x}};\text{w}'=\sec^2\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uvw})=\text{u}'\text{vw}+\text{uv}'\text{w}+\text{uv}\text{w}'$
$=\text{e}^{\text{x}}\log\sqrt{\text{x}}\tan\text{x}+\text{e}^\text{x}\times\frac{1}{\text{2x}}\tan\text{x}+\text{e}^\text{x}\log\sqrt{\text{x}}\sec^2\text{x}$
$=\text{e}^\text{x}(\log\text{x}^{\frac{1}{2}}.\tan\text{x}+\frac{\tan\text{x}}{\text{2x}}+\log\text{x}^{\frac{1}{2}}.\sec^2\text{x})$
$=\text{e}^\text{x}\Big(\frac{1}{2}\log\text{x}.\tan\text{x}+\frac{\tan\text{x}}{\text{2x}}+\frac{1}{2}\log\text{x}.\sec^2\text{x}\Big)$
$=\frac{\text{e}^\text{x}}{2}\Big(\log\text{x}.\tan\text{x}+\frac{\tan\text{x}}{\text{2x}}+\log\text{x}.\sec^2\text{x}\Big)$
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