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# Derviative of 1/u

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The derivative of 1/u for all u(x) not zero, is given by: -u’/u^2. Let’s prove it using limits.

## Derivative of 1/u(x)

Let $u(x)$ be a function of real variable $x$ such that $u(x) \neq 0$, and let $f(x) = \frac{1}{u(x)}$.

The derivative $f’(x)$ of the function $f(x)$ is:

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

## Proof

Let $x \in \mathbb{R}^*$ be such that $u(x) \neq 0$.

\begin{aligned} f’(x)=&\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\\ =&\lim_{h \rightarrow 0} \frac{\dfrac{1}{u(x+h)}-\dfrac{1}{u(x)}}{h}\\ =&\lim_{h \rightarrow 0} \frac{u(x)-u(x+h)}{h\cdot u(x)u(x+h)}\\ =&\lim_{h \rightarrow 0} -\frac{(u(x+h)-u(x))}{h\cdot u(x)u(x+h)}\\ =&\lim_{h \rightarrow 0} -\frac{(u(x+h)-u(x))}{h}\frac{1}{u(x)u(x+h)}\\ =&\lim_{h \rightarrow 0} -u’(x)\cdot\frac{1}{u(x)u(x+h)}\\ =&\lim_{h \rightarrow 0} -\frac{u’(x)}{u(x)u(x+h)}\\ =&-\frac{u’(x)}{u^2(x)} \end{aligned}

Therefore, we have:

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