Question

What is the derivative of ${{\log }_{4}}{{x}^{2}}$ ?

Answer 1

Hint : First we find the simplified form of ${{\log }_{4}}{{x}^{2}}$ using the formulas like $\log {{x}^{a}}=a\log x$ , ${{\log }_{b}}x=\dfrac{{{\log }_{c}}x}{{{\log }_{c}}b}$ . Then we find the derivative of $\dfrac{\ln x}{\ln 2}$ using $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$ . We find the final solution.

Complete step by step solution:
We first simplify the logarithm of ${{\log }_{4}}{{x}^{2}}$ . We apply the formula of ${{\log }_{b}}x=\dfrac{{{\log }_{c}}x}{{{\log }_{c}}b}$ .
For our given ${{\log }_{4}}{{x}^{2}}$ , we assume $c=e$ . We also know that $\ln a={{\log }_{e}}a$ .
Therefore, ${{\log }_{4}}{{x}^{2}}=\dfrac{{{\log }_{e}}{{x}^{2}}}{{{\log }_{e}}4}=\dfrac{{{\log }_{e}}{{x}^{2}}}{{{\log }_{e}}{{2}^{2}}}$
We have the properties where $\log {{x}^{a}}=a\log x$ . We apply that on both numerator and denominator.
${{\log }_{4}}{{x}^{2}}=\dfrac{{{\log }_{e}}{{x}^{2}}}{{{\log }_{e}}{{2}^{2}}}=\dfrac{2{{\log }_{e}}x}{2{{\log }_{e}}2}=\dfrac{{{\log }_{e}}x}{{{\log }_{e}}2}$
We now use the form of $\ln a={{\log }_{e}}a$ and get ${{\log }_{4}}{{x}^{2}}=\dfrac{{{\log }_{e}}x}{{{\log }_{e}}2}=\dfrac{\ln x}{\ln 2}$ .
The value of $\ln 2$ is a constant.
Now we find the derivative of ${{\log }_{4}}{{x}^{2}}$ being equal to $\dfrac{\ln x}{\ln 2}$ .
We know that the derivative form for $\ln x$ is $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$ .
Therefore, we get $\dfrac{d}{dx}\left( \dfrac{\ln x}{\ln 2} \right)=\dfrac{1}{\ln 2}\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x\ln 2}$ .
Therefore, the derivative of ${{\log }_{4}}{{x}^{2}}$ is $\dfrac{1}{x\ln 2}$ .
So, the correct answer is “ $\dfrac{1}{x\ln 2}$ ”.

Note : There are some particular rules that we follow in case of finding the condensed form of logarithm. We first apply the power property first. The chain rule may be extended or generalized to many other situations, including to products of multiple functions, to a rule for higher-order derivatives of a product, and to other contexts.