Derivative of 1/logx | 1 by log x Derivative

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 $-{\displaystyle \frac{1}{x(\log x{)}^2}}$.

Explanation:

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

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

${\displaystyle \frac{dz}{dx}}\log x+z{\displaystyle \frac{d}{dx}}(\log x)=0$ 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 ${\displaystyle \frac{dz}{dx}}\log x+z\cdot 1/x=0$

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

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

$\Rightarrow {\displaystyle \frac{dz}{dx}}=-{\displaystyle \frac{1}{x(\log x{)}^2}}$ 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:

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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 ${\displaystyle \frac{dz}{dx}}={\displaystyle \frac{1}{x}}$. Now by the chain rule of derivatives, we have that

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

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

$=-{\displaystyle \frac{1}{x{z}^2}}$

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

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

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