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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 $$y=u+v$$

$$\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$$

$$\dfrac{dy}{dx}=x^{x\cos x}.\left\{\cos x-x\sin x(\log x)+\cos x(\log x)\right\}+\dfrac{-4x}{(x^{2}-1)^{2}}$$

Mathematics
RS Agarwal
Standard XII

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