The derivative of 1/log x is -1/x(log x)^2. Note that 1/logx is the reciprocal of logx. In this post, we will learn how to find the derivative of 1 divided by log x.

## What is the Derivative of 1/logx?

**Answer:** The derivative of $1/\log x$ is $-\dfrac{1}{x(\log x)^2}$.

**Explanation:**

**Step 1:** Let $z = 1/\log x$. We need to find $\dfrac{dz}{dx}$.

**Step 2:** Cross multiplying, we obtain that

$z \log x =1$

**Step 3:** Differentiating both sides using the product rule with respect to x, we have that

$\dfrac{dz}{dx}\log x + z \dfrac{d}{dx}(\log x)=0$ as the derivative of a constant (here 1) is zero. For more details click the page on Derivative of a constant is 0).

**Step 4:** Simplify the above equation. We have $\dfrac{dz}{dx}\log x + z \cdot 1/x=0$

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

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

$\Rightarrow \dfrac{dz}{dx} = -\dfrac{1}{x(\log x)^2}$ as z=1/logx.

So the derivative of 1/logx is -1/x(logx)^2 and this is obtained by the product rule of derivatives.

**Also Read: **

Derivative of log(3x) | Derivative of log(sin x) |

Derivative of 1/x | Derivative of log(cos x) |

Derivative of 1/x^{2} | Derivative of xe^{-x} |

## Derivative of 1/logx by Chain Rule

To find the derivative of 1/logx using the chain rule, let us put z=log x. Note that $\dfrac{dz}{dx}=\dfrac{1}{x}$. Now by the chain rule of derivatives, we have that

$\dfrac{d}{dx}(\dfrac{1}{\log x})$ $=\dfrac{d}{dz}(\dfrac{1}{z}) \cdot \dfrac{dz}{dx}$

$=-\dfrac{1}{z^2} \cdot \dfrac{1}{x}$ as the derivative of 1/x is -1/x^2.

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

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

Thus, the derivative of 1/logx is equal to -1/x(log x)^2 obtained by the chain rule of derivatives.

**Have You Read These Derivatives?**

Derivative of sin^{2}x | Derivative of tanh(x) |

Derivative of tan2x | Derivative of tan^{2}x |

Derivative of sinh(x) | Derivative of cosh(x) |

## FAQs

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

Answer: The derivative of 1/logx is -1/x(logx)^2.