Solveeit Logo

Question

Question: How do you differentiate \[y={{\log }_{b}}x\]?...

How do you differentiate y=logbxy={{\log }_{b}}x?

Explanation

Solution

To solve this problem, we should know the properties of the logarithmic function and the derivative of the logarithmic function. We know the derivative of the function lnx\ln x with respect to x is 1x\dfrac{1}{x}. The logarithmic functions have a property by which we can change their bases as logba=logalogb{{\log }_{b}}a=\dfrac{\log a}{\log b}. We will use this property and the derivative of lnx\ln x to solve the given question.

Complete step by step answer:
We are asked to evaluate the derivative of the function y=logbxy={{\log }_{b}}x. We don’t know the direct derivative of this function, but we know that the derivative of the function lnx\ln x with respect to x is 1x\dfrac{1}{x}. The logarithmic functions have a property by which we can change their bases as logba=logalogb{{\log }_{b}}a=\dfrac{\log a}{\log b}.
We are given the function y=logbxy={{\log }_{b}}x. Changing the base of the logarithm, we get y=lnxlnby=\dfrac{\ln x}{\ln b}. As lnb\ln bis a constant, we can take it out while differentiating. Thus, we can evaluate derivative as,
dydx=1lnbd(lnx)dx\dfrac{dy}{dx}=\dfrac{1}{\ln b}\dfrac{d(\ln x)}{dx}. Substituting the derivative of the logarithmic function, we get

& \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln b}\dfrac{1}{x} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x\ln b} \\\ \end{aligned}$$ **Note:** We can use this question to make a general rule for these types of problems. Let’s say we are asked to differentiate the function of the form $$y={{\log }_{a}}x$$. Then, we can directly write the derivative of the functions as follows, $$\dfrac{dy}{dx}=\dfrac{1}{x\ln a}$$. Say we are given a function at the place of the argument of the logarithm, then in such cases do the same just multiply the derivative of the function in the argument at the end, that will be your final answer/ derivative of the function.