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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$(\tan\text{x})^\frac{1}{\text{x}}$
✓

Answer

Let $\text{y}=(\tan\text{x})^\frac{1}{\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\tan\text{x})^\frac{1}{\text{x}}$
$\log\text{y}=\frac{1}{\text{x}}\log(\tan\text{x})\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}\log(\tan\text{x})+\log(\tan\text{x})\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\Big)$
$=\frac{1}{\text{x}}\times\frac{1}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log(\tan\text{x})\Big(-\frac{1}{\text{x}^2}\Big)$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}\tan\text{x}}(\sec^2\text{x})-\frac{\log(\tan\text{x})}{\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sec^2\text{x}}{\text{x}\tan\text{x}}-\frac{\log(\tan\text{x})}{\text{x}^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^\frac{1}{\text{x}}\Big[\frac{\sec^2\text{x}}{\text{x}\tan\text{x}}-\frac{\log(\tan\text{x})}{\text{x}^2}\Big]$
[Using equation (i)]

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