Question.
$x+\frac{1}{x}^{3}$
$x+\frac{1}{x}^{3}$
solution:
Let $y=\left(x+\frac{1}{x}\right)^{3}$
Now differentiating y with respect to $x$ we get
$\Rightarrow$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{3}$
Expanding the equation using $(a+b)^{3}$ formula then we get
$=\frac{d}{d x}\left(x^{3}+\frac{1}{x^{3}}+3 x+\frac{3}{x}\right)$
Splitting the differential we get
$=\frac{d}{d x}\left(x^{3}\right)+\frac{d}{d x}\left(\frac{1}{x^{3}}\right)+\frac{d}{d x}(3 x)+\frac{d}{d x}\left(\frac{3}{x}\right)$
On differentiating we get
$=3 x^{2}-3 x^{-4}+3-3 x^{-2}$
$=3 x^{2}-\frac{3}{x^{4}}+3-\frac{3}{x^{2}}$
Let $y=\left(x+\frac{1}{x}\right)^{3}$
Now differentiating y with respect to $x$ we get
$\Rightarrow$$\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
$=3 x^{2}-3 x^{-4}+3-3 x^{-2}$
$=3 x^{2}-\frac{3}{x^{4}}+3-\frac{3}{x^{2}}$