How to find the derivative of 1/x. In this article, we will find the derivative of 1 divided by x using the power rule, product rule, and the definition of derivatives.

Table of Contents

**What is the Derivative of 1/x?**

At first, we will evaluate the derivative of 1/x by the power rule of derivatives. We need to follow the below steps.

Step 1: First, we will express 1/x as a power of x using the rule of indices. So we have

$1/x=x^{-1}$

Step 2: Now, we will apply the power rule of derivatives: $\frac{d}{dx}(x^n)=nx^{n-1}$. Thus we get that

$\frac{d}{dx}(1/x)=\frac{d}{dx}(x^{-1})=-1 \cdot x^{-1-1}$

Step 3: Simplifying the above expression, we obtain that

$\dfrac{d}{dx}(\dfrac{1}{x})=-1 \cdot x^{-2}$

$\Rightarrow \dfrac{d}{dx}(\dfrac{1}{x})=-1 \cdot \dfrac{1}{x^2}$

$\Rightarrow \dfrac{d}{dx}(\dfrac{1}{x})=\dfrac{-1}{x^2}$

So the derivative of $1/x$ is $-1/x^2$. In other words,

\[\dfrac{d}{dx}(\frac{1}{x})=-\dfrac{1}{x^2}.\]

**Also Read:**

**Derivative of e ^{sin x}** : The derivative of e

^{sin x }is cos x e

^{sin x}.

**Integration of mod x** : The integration of mod x is -x|x|/2+c

Now, we will find the derivative of 1/x by the first principle.

**Derivative of 1/x from first principle**

Let $f(x)=\dfrac{1}{x}.$ Applying the first principle of derivatives, we get that

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$

From the above definition of derivatives, the derivative of 1/x by first principle is equal to

$\dfrac{d}{dx}(\dfrac{1}{x})$ $= \lim\limits_{h \to 0} \dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}$

$=\lim\limits_{h \to 0}\dfrac{\frac{x-x-h}{x(x+h)}}{h}$

$=\lim\limits_{h \to 0}\dfrac{-h}{hx(x+h)}$

$=-\lim\limits_{h \to 0}\dfrac{1}{x(x+h)}$

$=-\dfrac{1}{x(x+0)}$

$=-\dfrac{1}{x^2}$

This shows that the formula of the derivative of 1/x is $-1/x^2$.

We know that the product rule of derivatives is $\frac{d}{dx}(fg)=f \frac{dg}{dx}+ g \frac{df}{dx}$. Using this rule, we will now find the derivative of 1/x.

**Derivative of 1/x by product rule**

Let us put $z=1/x$. This implies that

$zx=1$

Differentiating with respect to x, we get that

$\dfrac{d}{dx}(zx)=\dfrac{d}{dx}(1)$

$\Rightarrow z\dfrac{d}{dx}(x)+x\dfrac{d}{dx}(z)=0$ (by the product rule of derivatives)

$\Rightarrow z\cdot 1+x\dfrac{dz}{dx}=0$

$\Rightarrow x\dfrac{dz}{dx}=-z$

$\Rightarrow \dfrac{dz}{dx}=-\dfrac{z}{x}$

$\Rightarrow \dfrac{dz}{dx}=-\dfrac{1}{x^2}$ as z=1/x

So we have obtained the derivative of 1/x which is $-1/x^2$.

**Also Read: **

**Derivative of root x**: The derivative of √x is 1/2√x

**Derivative of cube root of x**: The derivative of the cube root of x is 1/(3x^{2/3})

**Derivative of sin inverse x**: The derivative of sin^{-1} x is 1/√(1-x^{2})

**Derivative of sin 3x**: The derivative of sin 3x is 3cos 3x

**Application of Derivative of 1/x**

As an application of the derivative of 1/x, we will now find the derivative of 1/log x. We will use the chain rule of derivatives: $\frac{du}{dx}=\frac{du}{dz} \cdot \frac{dz}{dx}$

**Question:** What is the derivative of $\frac{1}{\log x}$

**Solution:**

Let $z=\log x$. So we have $\frac{dz}{dx}=\frac{1}{x}$

Now, $\dfrac{d}{dx}(\dfrac{1}{\log x})$

$=\dfrac{d}{dx}(\dfrac{1}{z})$

$=\dfrac{d}{dz}(\dfrac{1}{z}) \cdot \dfrac{dz}{dx}$ (by the chain rule)

$=-\dfrac{1}{z^2} \cdot \dfrac{1}{x}$ (by the above formula of the derivative of 1/x)

$=-\dfrac{1}{x(\log x)^2}$ as z=log x.