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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\tan\text{x})^{\log\text{x}}+\cos^2\big(\frac{\pi}{4}\big)$
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Answer

Here,
$\text{y}=(\tan\text{x})^{\log\text{x}}+\cos^2\big(\frac{\pi}{4}\big)$
$\text{y}=\text{e}^{\log(\tan\text{x})^{\log\text{x}}}+\cos^2\big(\frac{\pi}{2}\big)$
$\text{y}=\text{e}^{\log\text{x}\log\tan\text{x}}+\cos\text{x}^2\big(\frac{\pi}{4}\big)$
$\big[\text{Since, e}^{\log\text{a}}=\text{a and}\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating ti using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\log\text{x}\log\tan\text{x}})+\frac{\text{d}}{\text{dx}}\cos^2\big(\frac{\pi}{4}\big)$
$=\text{e}^{\log\text{x}\log\tan\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x}\log\tan\text{x})+0$
$=\text{e}^{\log(\tan\text{x})^{\log\text{x}}}\Big[\log\times\frac{\text{d}}{\text{dx}}(\log\tan\text{x})+\log\tan\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{1}{\tan\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\text{x}\Big(\frac{1}{\text{x}}\Big)\Big]$
$=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{1}{\tan\text{x}}\Big)\big(\sec^2\text{x}\big)+\frac{\log\tan\text{x}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^{\log\text{x}}\Big[\log\times\Big(\frac{\sec^2\text{x}}{\tan\text{x}}\Big)+\frac{\log\tan\text{x}}{\text{x}}\Big]$

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