Gujarat BoardEnglish MediumSTD 11 ScienceMATHSDerivatives2 Marks
Question
Differentiate the following function with respect to $\text{x}$:$\text{x}^2\text{e}^\text{x}\log\text{x}$
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Answer
Let $\text{u}=\text{x}^2;\text{v}=\text{e}^\text{x};\text{w}=\log\text{x}$Then, $\text{u}'=2\text{x};\text{v}'=\text{e}^\text{x};\text{w}=\frac{1}{\text{x}}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uvw})=\text{u}'\text{vw}+\text{u}\text{v}'\text{w}+\text{uv}\text{w}'$
$\frac{\text{d}}{\text{dx}}(\text{x}^2\text{e}^\text{x}\log\text{x})=\text{2xe}^\text{x}\log\text{x}+\text{x}^2\text{e}^\text{x}\log\text{x}+\text{x}^2\text{e}^\text{x}\frac{1}{\text{x}}$
$=\text{2xe}^\text{x}\log\text{x}+\text{x}^2\text{e}^\text{x}\log\text{x}+\text{x}\text{e}^\text{x}$
$=\text{x}\text{e}^\text{x}(2\log\text{x}+\text{x}\log\text{x}+1)$
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