Find the derivative of $f(x) = \dfrac{x + 1}{x}$.

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> [!TLDR]
> The derivative of $f(x) = \frac{x+1}{x}$ is found by rewriting the function as $1 + x^{-1}$ and then differentiating term by term, giving $f'(x) = -\frac{1}{x^2}$.

The key insight here is that **derivative at a point** measures the instantaneous rate of change — the slope of the tangent line. But to find the derivative *function*, we need a rule that works for every $x$ in the domain. The quotient rule is one option, but it's often messier than necessary. A cleaner path: simplify the algebraic form first.

When you see a rational function like $\frac{x+1}{x}$, ask: can I split it into simpler pieces? Yes — because the numerator is a sum, you can divide term by term:

$$f(x) = \frac{x}{x} + \frac{1}{x} = 1 + \frac{1}{x}$$

This is much friendlier. Now $f(x) = 1 + x^{-1}$, and we know the derivative of a constant is zero, and the power rule gives $\frac{d}{dx} x^n = n x^{n-1}$.

Let's walk through it step by step.

1. **Rewrite the function**  
   $$f(x) = \frac{x+1}{x} = 1 + \frac{1}{x} = 1 + x^{-1}$$  
   This works for all $x 
eq 0$, which is the domain anyway.

2. **Differentiate term by term**  
   The derivative of the constant $1$ is $0$.  
   For $x^{-1}$, apply the power rule with $n = -1$:  
   $$\frac{d}{dx} x^{-1} = (-1) x^{-2} = -\frac{1}{x^2}$$

3. **Combine the results**  
   $$f'(x) = 0 + \left(-\frac{1}{x^2}ight) = -\frac{1}{x^2}$$ …

Q.Find the derivative of $f(x) = \dfrac{x + 1}{x}$ .

Concept understanding — Rules of Differentiation

Q.Find the derivative of f(x)=x+1xf(x) = \dfrac{x + 1}{x}f(x)=xx+1​.

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Q.Find the derivative of $f(x) = \dfrac{x + 1}{x}$ .

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