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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $(\cos\text{x})^{\text{y}}=(\tan\text{y})^{\text{x}},$ Prove that $\frac{\text{dy}}{\text{dx}}=\frac{\log\tan\text{y}-\text{y}\tan\text{x}}{\log\cos\text{x}-\text{x}\sec\text{y cosec y}}$
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Answer

Here,
$(\cos\text{x})^{\text{y}}=(\tan\text{y})^{\text{x}}$
Taking log on both sides,
$\log(\cos\text{x})^{\text{y}}=\log(\tan\text{y})^{\text{x}}$
$\text{y}\log(\cos\text{x})=\text{x}\log(\tan\text{y})$
$\big[\text{Since}, \log\text{e}^{\text{b}}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule and product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log\cos\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}\log\tan\text{y})$
$\Big(\text{y}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ =\Big(\text{x}\frac{\text{d}}{\text{dx}}\log\tan\text{y}+\log\tan\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big)$
$\Big(\text{y}\big(\frac{1}{\cos\text{x}}\big)\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ =\Big(\text{x}\frac{1}{\tan\text{y}}\frac{\text{d}}{\text{dx}}(\tan\text{y})+\log\tan\text{y}(1)\Big)$
$\Big(\frac{\text{y}}{\cos\text{x}}(-\sin\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}\Big)\\ =\Big(\frac{\text{x}}{\tan\text{y}}(\sec^2\text{y})\Big)\frac{\text{dy}}{\text{dx}}+\log\tan\text{y}-\text{y}\tan\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}} \\ =\Big(\sec\text{y cosec y}\times\text{y}\frac{\text{dy}}{\text{dx}}+\log\tan\text{y}\Big)$
$\frac{\text{dy}}{\text{dx}}\big[\log\cos\text{x}-\text{x}\sec\text{y cosec y}\big] \\ =\log\tan\text{y}+\text{y}\tan\text{x}$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{\log\tan\text{x}+\text{y}\tan\text{x}}{\log\cos\text{x}-\text{x}\sec\text{y cosec y}}\Big]$

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