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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
If `$\text{y}=(\cos\text{x})^{(\cos\text{x})^{(\cos\text{x})\dots\infty}},$ show that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2\tan\text{x}}{\text{y}\log\cos\text{x}-1}.$
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Answer

We have, $\text{y}=(\cos\text{x})^{(\cos\text{x})^{(\cos\text{x})\dots\infty}}$
$\Rightarrow\ \text{y}=(\cos\text{x})^\text{y}$
$\therefore\ \log\text{y}=\log(\cos\text{x})^\text{y}$
Differentiating w.r.t. x, we get
$\frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\text{y}\cdot\frac{\text{d}}{\text{dx}}(\log\cos\text{x})+\log\cos\text{x}\cdot\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\cos\text{x}}\cdot\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\cdot\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big[\frac{1}{\text{y}}-\log\cos\text{x}\Big]$ $=\frac{-\text{y}\sin\text{x}}{\cos\text{x}}=-\text{y}\tan\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2\tan\text{x}}{(1-\text{y}\log\cos\text{x})}=\frac{\text{y}^2\tan\text{x}}{\text{y}\log\cos\text{x}-1}$

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