$x+\frac{1}{x}^{3}$
Question.

$x+\frac{1}{x}^{3}$

solution:

Let $y=\left(x+\frac{1}{x}\right)^{3}$

Now differentiating y with respect to $x$ we get

$\Rightarrow$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{3}$

Expanding the equation using $(a+b)^{3}$ formula then we get

$=\frac{d}{d x}\left(x^{3}+\frac{1}{x^{3}}+3 x+\frac{3}{x}\right)$

Splitting the differential we get

$=\frac{d}{d x}\left(x^{3}\right)+\frac{d}{d x}\left(\frac{1}{x^{3}}\right)+\frac{d}{d x}(3 x)+\frac{d}{d x}\left(\frac{3}{x}\right)$

On differentiating we get

$=3 x^{2}-3 x^{-4}+3-3 x^{-2}$

$=3 x^{2}-\frac{3}{x^{4}}+3-\frac{3}{x^{2}}$
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