Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 Marks
Question
Differentiate the following functions with respect to x: $(\log\text{x})^\text{x}$
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Answer
Let $\text{y}=(\log\text{x})^\text{x}\ .....(\text{i})$ Taking log on both the sides, $\log\text{y}=\log(\log\text{x})^\text{x}$ $\Rightarrow\log\text{y}=\text{x}\log(\log\text{x})\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$ Differentiating with respect to x, using product rule, chain rule, $\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log(\log\text{x})+\log\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})$ $=\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\log\text{x}(1)$ $=\frac{\text{x}}{\log\text{x}}\Big(\frac{1}{\text{x}}\Big)+\log\log\text{x}$ $\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\log\text{x}}+\log\log\text{x}$ $\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]$ $\frac{\text{dy}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log\log\text{x}\Big]$ [Using equation (i)]
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