Question

Use implicit differentiation and the fact that ${\log }_{b}x$ is the inverse of $b^x$ to prove that$\frac{d}{dx}\left({\log }_{b}x\right)=\frac{1}{x\ln b}$

Short answer

The given differentiation has been proved.

Step-by-step solution

Step 1. Given Information   

The given function is${\log }_{b}x$

Step 2. Proof   

Let ${\log }_{b}x=y\Rightarrow x=b^y$

Take log on both the sides, we get,

$$\begin{gathered} \ln x=\ln b^y \\ \ln x=y\ln b \\ \frac{1}{x}=\ln b\frac{dy}{dx} \\ \frac{dy}{dx}=\frac{1}{x\ln b} \\ \frac{d}{dx}\left({\log }_{b}x\right)=\frac{1}{xlnb} \end{gathered}$$

Hence, Proved.