Question

# Find $$\dfrac {dy}{dx}$$, when $$y=x^{x\cos x} +\left (\dfrac {x^2 +1}{x^2 -1}\right)$$.

Solution

## Given, $$y=x^{x\cos x} +\left (\dfrac {x^2 +1}{x^2 -1}\right)$$Let $$u=x^{x\cos x}$$ and $$v=\left(\dfrac{x^{2}+1}{x^{2}-1}\right)$$. Then,Now, $$u=x^{x\cos x}$$ Taking log on both side$$\log u=(x\cos x)\log x$$Differentiate with respect to $$x$$.$$\Rightarrow \dfrac{1}{u}.\dfrac{du}{dx}=(x\cos x).\dfrac{1}{x}+(\log x)(-x\sin x+\cos x)\\$$$$\Rightarrow \dfrac{du}{dx}=u.\left\{\cos x-x\sin x(\log x)+\cos x(\log x)\right\}\\$$$$\Rightarrow \dfrac{du}{dx}=x^{x\cos x}.\left\{\cos x-x\sin x(\log x)+\cos x(\log x)\right\}$$Now,$$v=\left(\dfrac{x^{2}+1}{x^{2}-1}\right)$$.Differentiate with respect to $$x$$.$$\dfrac{dv}{dx}=\dfrac{(x^{2}-1).2x-(x^{2}+1).2x}{(x^{2}-1)^{2}}$$$$=\dfrac{-4x}{(x^{2}-1)^{2}}$$As $$y=u+v$$$$\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$$$$\dfrac{dy}{dx}=x^{x\cos x}.\left\{\cos x-x\sin x(\log x)+\cos x(\log x)\right\}+\dfrac{-4x}{(x^{2}-1)^{2}}$$MathematicsRS AgarwalStandard XII

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