Question
Differentiate $x^{x \cos x}$ + $\frac{\text{x}^{2}+1}{\text{x}^{2}-1}$ w.r.t.x.

Answer

$​​\text{y}=​\text{x}^{\text{x cos x}}+​\frac{\text{x}^{2}+1}{\text{x}^{2}-1}=\text{u + v}$$\text{u}=\text{x}^{\text{x cos x}}\Rightarrow\log\text{u}=\text{x cos x}\cdot{\text{log x}}$
$\therefore\frac{1}{\text{u}}\cdot\frac{\text{du}}{\text{dx}}=\frac{\text{x cos x}}{\text{x}}+\text{cos x}\cdot{\text{log x - sin x log x}}$
$\Rightarrow$$\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x cos x}}(\text{cos x+cos x log x}\text{ - sin x log x})$
$\text{v}=\frac{\text{x}^{2}+1}{\text{x}^{2}-1},\frac{\text{dv}}{\text{dx}}=\frac{(\text{x}^{2}-1)\text{2x}-\text{(x}^{2}+1)\text{2x}}{\text{(x}^{2}-1)^{2}}=-\frac{\text{4x}}{\text{(x}^{2}-1)^{2}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x cos x}}\text{(cos x + log x}\cdot\text{cos x - sin x log x})-\frac{\text{4x}}{(\text{x}^{2}-1)^{2}}$

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