Question

Differentiate

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

Answer 1

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

$\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

$\therefore \frac{dy}{dx}=\frac{d}{dx}\left(\frac{x^4}{4}\right)-\frac{d}{dx}\left(\frac{x^{-3}}{3}\right)-\frac{d}{dx}\left(\frac{2}{x}\right)+0$ ($\because C$ is constant)

$=\frac{4x^3}{4}-\frac{-3x^{-4}}{3}-2(-1)x^{-2}$

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

$\therefore \frac{dy}{dx}=x^3+\frac{1}{x^4}+\frac{2}{x^2}$