The derivative of 1/x is equal to -1/x^{2}. The function 1/x (1 divided by x) is the reciprocal of x. In this article, we will find the derivative of 1/x using the power rule, product rule, and the definition of derivatives.

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

Answer: The derivative of 1/x is denoted by d/dx (1/x) and it is equal to -1/x^{2}. |

**Explanation:**

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

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

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

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

So the derivative of 1/x by the power rule is -1/x^{2} and this is obtained by the power rule of derivatives.

**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)=1/x. Applying the first principle of derivatives, we get that

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

Thus, the derivative of 1/x by first principle is equal to

$\dfrac{d}{dx}(\dfrac{1}{x})$ = lim_{h→0} $\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}$

= lim_{h→0} $\dfrac{\frac{x-x-h}{x(x+h)}}{h}$

= lim_{h→0} $\dfrac{-h}{hx(x+h)}$

= – lim_{h→0} $\dfrac{1}{x(x+h)}$

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

= -1/x^{2}

This shows that the derivative of 1/x is -1/x^{2} and this is obtained by the first principle of derivatives.

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

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

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

⇒ $z\cdot 1+x\dfrac{dz}{dx}=0$

⇒ $x\dfrac{dz}{dx}=-z$

⇒ $\dfrac{dz}{dx}=-\dfrac{z}{x}$

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

So the derivative of 1/x by the product rule is equal to -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 1/logx.

**Solution:**

Let z=log x. So we have dz/dx=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.

## FAQs

**Q1: What is the derivative of 1/x?**

Answer: The derivative of 1/x is -1/x^{2}.

**Q2: What is the antiderivative of 1/x?**

Answer: The anti-derivative of 1/x is ln x.