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Differentiation — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentiation5 Marks
Question
If $\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}},$ prove that $\frac{\text{y}^2\cot\text{x}}{(1-\text{y}\log\sin\text{x})}$
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Answer

Here,
$\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}}$
$\Rightarrow\text{y}=(\sin\text{x})^\text{y}$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x})^{\text{y}}$
$\log\text{y}=\text{y}(\log\sin\text{x})$
Differentiating it with respect to x, using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\sin\text{x}\Big)=\frac{\text{y}}{\sin\text{x}}(\cot\text{x})$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1-\text{y}\log\sin\text{x}}{\text{y}}\Big)=\text{y}\cot\text{x}$
$\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{y}^2\cot\text{x}}{1-\text{y}\log\sin\text{x}}\Big)$

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