Question

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

Solution

Let $$\displaystyle y = x^{x \cos x} + \frac{x^2 + 1}{x^2 - 1}$$Also, let $$\displaystyle u = x^{x \cos x}$$ and $$\displaystyle v = \frac{x^2 + 1}{x^2 - 1}$$$$\therefore y = u + v$$$$\Rightarrow \displaystyle \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$$             ...(1)$$\displaystyle u = x^{x \cos x}$$$$\Rightarrow \displaystyle \log u = \log (x^{x \cos x})$$$$\Rightarrow \displaystyle \log u = x \cos x \log x$$Differentiating both sides with respect to $$x$$, we obtain$$\displaystyle \frac{1}{u} \frac{du}{dx} = \frac{d}{dx} (x) . \cos x . \log x + x \frac{d}{dx} (\cos x) . \log x + x \cos x . \frac{d}{dx} (\log x)$$$$\Rightarrow \displaystyle \frac{du}{dx} = u \left[ 1 . \cos x . \log x + x . (- \sin x) \log x + x \cos x . \frac{1}{x} \right]$$$$\Rightarrow \displaystyle \frac{du}{dx} = x^{x \cos x} \left( \cos x . \log x - x \sin x \log x + \cos x \right)$$$$\Rightarrow \displaystyle \frac{du}{dx} = x^{x \cos x} \left[ \cos x (1 + \log x) - x \sin x \log x \right]$$                                    ...(2)$$\displaystyle v = \frac{x^2 + 1}{x^2 - 1}$$$$\Rightarrow \displaystyle \log v = \log {(x^2 + 1)} - \log {(x^2 - 1)}$$Differentiating both sides with respect to $$x$$, we obtain$$\displaystyle \frac{1}{v} \frac{dv}{dx} = \frac{2x}{x^2 + 1} - \frac{2x}{x^2 - 1}$$$$\Rightarrow \displaystyle \frac{dv}{dx} = v \left[ \frac{2x(x^2 - 1) - 2x(x^2 + 1)}{(x^2 + 1) (x^2 - 1)} \right]$$$$\Rightarrow \displaystyle \frac{dv}{dx} = \frac{x^2 + 1}{x^2 - 1} \times \left[ \frac{- 4x }{(x^2 + 1) (x^2 - 1)} \right]$$$$\Rightarrow \displaystyle \frac{dv}{dx} = \frac{- 4x }{(x^2 - 1)^2}$$                        ...(3)From (1), (2) and (3), we obtain$$\Rightarrow \displaystyle \frac{dy}{dx} = x^{x \cos x} \left[ \cos x (1 + \log x) - x \sin x \log x \right] - \frac{4x }{(x^2 - 1)^2}$$MathematicsRS AgarwalStandard XII

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