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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the functions given in Exercise:
$\text{x}^{\sin\text{x}}+(\sin\text{x})^{\cos\text{x}}$
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Answer

Let $\text{y}=\text{x}^{\sin\text{x}}+(\sin\text{x})^{\cos\text{x}}$
Putting $\text{u}=\text{x}^{\sin\text{x}}\text{ and v }(\sin\text{x})^{\cos\text{x}},\text{we get }\text{ y}=\text{u}+\text{v}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now $\text{u}=\text{x}^{\sin\text{x}}\ \Rightarrow\ \log\text{u}=\log\text{x}^{\sin\text{x}}=\sin\text{x}\log\text{x}$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{u}=\frac{\text{d}}{\text{dx}}(\sin\text{x}\log\text{x})$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\sin\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}\sin\text{x}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\sin\text{x}\frac{1}{\text{x}}+\log\text{x}(\cos\text{x})$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}\Big(\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}\Big)$ $\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big(\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}\Big) \dots\text{(ii)}$
Again $\text{v}=(\sin\text{x})^{\cos\text{x}}\ \Rightarrow\ \log\text{v}=\log(\sin\text{x})^{\cos\text{x}}=\cos\text{x}\log\sin\text{x}$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{v}=\frac{\text{d}}{\text{dx}}[\cos\text{x}\log(\sin\text{x})]$ $\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}\cos\text{x}$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\cos\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}\sin\text{x}+\log\sin\text{x}(-\sin\text{x})$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\cot\text{x}.\cos\text{x}-\sin\text{x}\log\sin\text{x}$ $\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{v}(\cot\text{x}.\cos\text{x}-\sin\text{x}\log\sin\text{x})$
$ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=(\sin\text{x})^{\cos\text{x}}(\cot\text{x}.\cos\text{x}-\sin\text{x}\log\sin\text{x})\ \dots\text{(iii)}$
Putting values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\sin\text{x}}\Big(\frac{\sin\text{x}}{\text{x}}+\cos\text{x}\log\text{x}\Big)+(\sin\text{x})^{\cos\text{x}}(\cot\text{x}.\cos\text{x}-\sin\text{x}\log\sin\text{x})$

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