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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\sin\text{x})^{\cos\text{x}}+(\cos\text{x})^{\sin\text{x}}$
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Answer

We have, $\text{y}=(\sin\text{x})^{\cos\text{x}}+(\cos\text{x})^{\sin\text{x}}$
$\Rightarrow\text{y}=\text{e}^{\log(\sin\text{x})^{\cos\text{x}}}+\text{e}^{\log(\cos\text{x})^{\sin\text{x}}}$
$\Rightarrow\text{y}=\text{e}^{\cos\text{x}\log\sin\text{x}}+\text{e}^{\sin\text{x}\log\cos\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}\log\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sin\text{x}\log\cos\text{x}}\big)$
$=\text{e}^{\cos\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}\big(\cos\text{x}\log\sin\text{x})+\text{e}^{\sin\text{x}\log\cos\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x}\log\cos\text{x})$
$=\text{e}^{\log(\sin\text{x})^{\cos\text{x}}}\Big[\cos\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})\Big] \\ +\text{e}^{\log(\cos\text{x})^{\sin\text{x}}}\Big[\sin\text{x}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})\Big]$
$=(\sin\text{x})^{\cos\text{x}}\Big[\cos\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\times(-\sin\text{x})\Big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\sin\text{x}\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x})+\log\cos\text{x}\times(\cos\text{x})\Big]$
$=(\sin\text{x})^{\cos\text{x}}\big[\cot\text{x}\cos\text{x}-\sin\text{x}\log\sin\text{x}\big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\tan\text{x}(-\sin\text{x})+\cos\text{x}\log\cos\text{x}\big]$
$=(\sin\text{x})^{\cos\text{x}}\big[\cot\text{x}\cos\text{x}-\sin\text{x}\log\sin\text{x}\big] \\ +(\cos\text{x})^{\sin\text{x}}\big[\cos\text{x}\log\cos\text{x}-\sin\text{x}\tan\text{x}\big]$

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