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Derivative of 1/x

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The derivative f’ of the function f(x)=1/x is: f’(x) = -1/x^2 for all nonzero real numbers x

Derivative of 1/x

The derivative $f’$ of the function $f(x)=\dfrac{1}{x}$ is:

$$ \forall x \in \mathbb{R}^* , f’(x) = -\frac{1}{x^2}$$

Proof/Demonstration

Let $x \in \mathbb{R}^*$

$$ \begin{aligned} \frac{df}{dx}=&\lim_{h \rightarrow 0} \frac{\displaystyle\frac{1}{x+h}-\frac{1}{x}}{h}\\ =&\lim_{h \rightarrow 0} \frac{\displaystyle \frac{1}{x+h}\cdot \frac{x}{x}- \frac{1}{x}\cdot \frac{x+h}{x+h}}{h}\\ =&\lim_{h \rightarrow 0} \frac{\displaystyle\frac{x-(x+h)}{x(x+h)}}{h}\\ =&\lim_{h \rightarrow 0} \frac{\displaystyle\frac{-h}{x(x+h)}}{h}\\ =&\lim_{h \rightarrow 0} \frac{\displaystyle\frac{-1}{x(x+h)}}{1}\\ =&\lim_{h \rightarrow 0} \frac{-1}{x(x+h)}=\frac{-1}{x(x+0)}\\ =&-\frac{1}{x^{2}} \end{aligned} $$

Then we have:

$$ \forall x \in \mathbb{R}^* , f’(x) = -\frac{1}{x^2}$$

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