Question

Differentiate the function given below w.r.t. $x:$

$\frac{x^4+x^3+x^2+1}{x}$

Answer 1

Let $y=\frac{x^4+x^3+x^2+1}{x}$

$\frac{dy}{dx}=\frac{d}{dx}(\frac{x^4}{x}+\frac{x^3}{x}+\frac{x^2}{x}+\frac{1}{x})$

$\frac{dy}{dx}=\frac{d}{dx}(x^3+x^2+x+\frac{1}{x})$

$\frac{dy}{dx}=\frac{d\left(x^3\right)}{dx}+\frac{d\left(x^2\right)}{dx}+\frac{d\left(x\right)}{dx}+\frac{d\left(\frac{1}{x}\right)}{dx}$

$[\because \frac{d(f+g)}{dx}=\frac{df}{dx}+\frac{dg}{dx}]$

$\frac{dy}{dx}=3x^2+2x+1-\frac{1}{x^2}$ $[\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}]$