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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{y}=(\cos\text{x})^{\cos\text{x}^{\cos\text{x}^{.....\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^2\tan\text{x}}{(1-\text{y}\log\cos\text{x})}$
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Answer

Here,
$\text{y}=(\cos\text{x})^{\cos\text{x}^{\cos\text{x}^{.....\infty}}}$
$\text{y}=(\cos\text{x})^\text{y}$
Taking log on both the sides,
$\log\text{y}=\log(\cos\text{x})^\text{y}$
$\log\text{y}=\text{y}\log(\cos\text{x}),\big\{\text{Since},\log\text{a}^\text{b}=\text{b}\log\text{a}\big\}$
Differentiating it with resepect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}\log(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\Big(\frac{1}{\cos\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\cos\text{x}\Big)=\frac{\text{y}}{\cos\text{x}}(-\sin\text{x})$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1-\log\cos\text{x}}{\text{y}}\Big)=-\text{y}\tan\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2\tan\text{x}}{(1-\log\cos\text{x})}$

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