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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $(\sin\text{x})^{\text{y}}=\text{x}+\text{y},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1-(\text{x}+\text{y})\text{y}\cot\text{x}}{(\text{x}+\text{y})\log\sin\text{x}-1}$
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Answer

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

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