Question.
$\frac{x^{4}+x^{3}+x^{2}+1}{x}$
$\frac{x^{4}+x^{3}+x^{2}+1}{x}$
solution:
Let $y=\frac{x^{4}+x^{3}+x^{2}+1}{x}$
$\Rightarrow$$y=\frac{x^{4}+x^{3}+x^{2}+1}{x}$
Dividing by $x$ we get
$\Rightarrow$$y=x^{3}+x^{2}+x+\frac{1}{x}$
Differentiating given equation with respect to $x$
$\Rightarrow$$\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}+x^{2}+x+\frac{1}{x}\right)$
On differentiation we get
$\Rightarrow$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}+\mathrm{x}^{2}+\mathrm{x}+\frac{1}{\mathrm{x}}\right)=3 \mathrm{x}^{2}+2 \mathrm{x}+1-\frac{1}{\mathrm{x}^{2}}$
$\Rightarrow$$\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}+x^{2}+x+\frac{1}{x}\right)$
On differentiation we get
$\Rightarrow$$\frac{d}{d x}\left(x^{3}+x^{2}+x+\frac{1}{x}\right)=3 x^{2}+2 x+1-\frac{1}{x^{2}}$
Hence, the required answer is $3 \mathrm{x}^{2}+2 \mathrm{x}+1-1 / \mathrm{x}^{2}$
Let $y=\frac{x^{4}+x^{3}+x^{2}+1}{x}$
$\Rightarrow$$y=\frac{x^{4}+x^{3}+x^{2}+1}{x}$
Dividing by $x$ we get
$\Rightarrow$$y=x^{3}+x^{2}+x+\frac{1}{x}$
Differentiating given equation with respect to $x$
$\Rightarrow$$\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}+x^{2}+x+\frac{1}{x}\right)$
On differentiation we get
$\Rightarrow$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}+\mathrm{x}^{2}+\mathrm{x}+\frac{1}{\mathrm{x}}\right)=3 \mathrm{x}^{2}+2 \mathrm{x}+1-\frac{1}{\mathrm{x}^{2}}$
$\Rightarrow$$\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}+x^{2}+x+\frac{1}{x}\right)$
On differentiation we get
$\Rightarrow$$\frac{d}{d x}\left(x^{3}+x^{2}+x+\frac{1}{x}\right)=3 x^{2}+2 x+1-\frac{1}{x^{2}}$
Hence, the required answer is $3 \mathrm{x}^{2}+2 \mathrm{x}+1-1 / \mathrm{x}^{2}$