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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{y}=(\tan\text{x})^{(\tan\text{x})^{(\tan\text{x})^{....\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=2\text{ at x}=\frac{\pi}{4}$
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Answer

We have, $\text{y}=(\tan\text{x})^{(\tan\text{x})^{(\tan\text{x})^{....\infty}}}$
$\Rightarrow\text{y}=(\tan\text{x})^{\text{y}}$
Taking log on both sides,
$\log\text{y}=\log(\tan\text{x})^\text{y}$
$\Rightarrow\log\text{y}=\text{y}\log\tan\text{x}$
Differentaiting with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}\big\{\log\tan\text{x}\big\}+\log\tan\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\frac{\text{d}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\tan\text{x}\Big)=\frac{\text{y}}{\tan\text{y}}\sec^2\text{x}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\text{y}\sec^3\big(\frac{\pi}{4}\big)}{\tan\big(\frac{\pi}{4}\big)}\times\frac{\text{y}}{1-\text{y}\log\tan\big(\frac{\pi}{4}\big)}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\text{y}^2\big(\sqrt{2}\big)^2}{1(1-\text{y}\log\tan1)}$
$\Rightarrow \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{2(1)^2}{(1-0)}\Bigg[\because(\text{y})_{\frac{\pi}{4}}=\big(\tan\frac{\pi}{4}\big)^{\big(\tan\frac{\pi}{4}\big)^{\big(\tan\frac{\pi}{4}\big)^{\ .....\infty}}}=1\Bigg]$
$\Rightarrow \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=2$

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