Derviative of 1/u

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 $u(x)\ne 0$, and let $f(x)=\frac{1}{u(x)}$.

The derivative $f^{\prime }(x)$ of the function $f(x)$ is: $\mathrm{\forall }x\in {\mathbb{R}}^{\ast },f^{\prime }(x)=-{\displaystyle \frac{u^{\prime }(x)}{u^2(x)}}$

Proof

Let $x\in {\mathbb{R}}^{\ast }$ be such that $u(x)\ne 0$. $\begin{array}{rl}{f}^{\prime }(x)=& \underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ =& \underset{h\to 0}{lim}\frac{{\displaystyle \frac{1}{u(x+h)}}-{\displaystyle \frac{1}{u(x)}}}{h}\\ =& \underset{h\to 0}{lim}\frac{u(x)-u(x+h)}{h\cdot u(x)u(x+h)}\\ =& \underset{h\to 0}{lim}-\frac{(u(x+h)-u(x))}{h\cdot u(x)u(x+h)}\\ =& \underset{h\to 0}{lim}-\frac{(u(x+h)-u(x))}{h}\frac{1}{u(x)u(x+h)}\\ =& \underset{h\to 0}{lim}-u^{\prime }(x)\cdot \frac{1}{u(x)u(x+h)}\\ =& \underset{h\to 0}{lim}-\frac{u^{\prime }(x)}{u(x)u(x+h)}\\ =& -\frac{u^{\prime }(x)}{u^2(x)}\end{array}$

Therefore, we have: $\mathrm{\forall }x\in {\mathbb{R}}^{\ast },f^{\prime }(x)=-{\displaystyle \frac{u^{\prime }(x)}{u^2(x)}}$