Question

differentiate x + 1/x from first principle

Answer 1

$$\begin{gathered} \mathrm{Let}y=x+\frac{1}{x} \\ \Rightarrow y+∆y=\left(x+∆x\right)+\frac{1}{\left(x+∆x\right)} \\ \Rightarrow ∆y=\left(x+∆x\right)+\frac{1}{\left(x+∆x\right)}-y \\ \Rightarrow ∆y=\left(x+∆x\right)+\frac{1}{\left(x+∆x\right)}-x-\frac{1}{x} \\ \Rightarrow ∆y=∆x+\frac{1}{\left(x+∆x\right)}-\frac{1}{x} \\ \Rightarrow ∆y=∆x+\left[\frac{x-x-∆x}{x\left(x+∆x\right)}\right] \\ \Rightarrow \frac{∆y}{∆x}=\frac{∆x}{∆x}-\left[\frac{∆x}{∆x}\times \frac{1}{x\left(x+∆x\right)}\right] \\ \Rightarrow \frac{∆y}{∆x}=1-\frac{1}{x\left(x+∆x\right)} \\ \Rightarrow \underset{∆x\to 0}{\lim }\frac{∆y}{∆x}=\underset{∆x\to 0}{\lim }\left[1-\frac{1}{x\left(x+∆x\right)}\right] \\ \Rightarrow \boxed{\frac{dy}{dx}=1-\frac{1}{x^2}} \end{gathered}$$