Differentiate x^ x x + x^ 2 +1 x^ 2 -1 w.r.t.x.

Question

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

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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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