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

Solution

$Let y = x^{x \cos x} and z = \frac{x^{2} + 1}{x^{2} - 1} Consider y = x^{x \cos x}$

Taking log on both sides,

log y = log ( x^{x co}^{s x} )

log y = x cos x log x

Differentiating with respect to x,

$\frac{1}{y} \frac{dy}{dx} = (x \cos x) \frac{1}{x} + \log x \frac{d}{dx} (x \cos x) \frac{1}{y} \frac{dy}{dx} = \cos x + \log x (\cos x - x \sin x) \frac{dy}{dx} = y [\cos x + \log x (\cos x - x \sin x)] \frac{dy}{dx} = x^{x \cos x} [\cos x + \log x (\cos x - x \sin x)] . . . . . . . (i) Consider z = \frac{x^{2} + 1}{x^{2} - 1}$

Differentiating with respect to x,

$\frac{dz}{dx} = \frac{(x^{2} - 1) . \frac{d}{dx} (x^{2} + 1) - (x^{2} + 1) . \frac{d}{dx} (x^{2} - 1)}{{(x^{2} + 1)}^{2}} = \frac{(x^{2} - 1) (2 x) - (x^{2} + 1) (2 x)}{{(x^{2} + 1)}^{2}} = \frac{2 x^{3} - 2 x - 2 x^{3} - 2 x}{{(x^{2} + 1)}^{2}} = \frac{- 4 x}{{(x^{2} + 1)}^{2}} . . . . . . . . . . . . (ii)$

Adding (i) and (ii)

$\frac{d}{dx} \{x^{x \cos x} + \frac{x^{2} + 1}{x^{2} - 1}\} = \frac{dy}{dx} + \frac{dz}{dx} = x^{x \cos x} [\cos x + \log x (\cos x - x \sin x)] - \frac{4 x}{{(x^{2} - 1)}^{2}}$

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