Question

What is the derivative of ${{\log }_{2}}\dfrac{{{x}^{2}}}{x-1}?$

Answer 1

We know the logarithmic identity ${{\log }_{a}}\dfrac{x}{y}={{\log }_{a}}x-{{\log }_{a}}y.$ We will use the logarithmic identity ${{\log }_{a}}{{x}^{n}}=n{{\log }_{a}}x.$ Then we will convert the logarithm to the base $a$ into natural logarithm by using the identity ${{\log }_{a}}x=\dfrac{\ln x}{\ln a}.$ And then, we will differentiate the simplified form we have obtained.

Complete step by step solution:
Let us consider the given logarithm to the base $2$ function ${{\log }_{2}}\dfrac{{{x}^{2}}}{x-1}$
We are asked to find the derivative of the given logarithmic function.
Now, we will use the logarithmic identity given by ${{\log }_{a}}\dfrac{x}{y}={{\log }_{a}}x-{{\log }_{a}}y.$
If we use the above identity, we will get the given function as ${{\log }_{2}}\dfrac{{{x}^{2}}}{x-1}={{\log }_{2}}{{x}^{2}}-{{\log }_{2}}\left( x-1 \right).$
Now, we can use the logarithmic identity given by ${{\log }_{a}}{{x}^{n}}=n{{\log }_{a}}x.$
Then, we will get ${{\log }_{2}}\dfrac{{{x}^{2}}}{x-1}=2{{\log }_{2}}x-{{\log }_{2}}\left( x-1 \right).$
Now, we can change the logarithm to the base $a$ into the natural logarithm using the logarithmic identity given by ${{\log }_{a}}x=\dfrac{\ln x}{\ln a}.$
When we use this identity, we can change the above obtained simplified form of the function which is in terms of logarithm to the base $2$ in terms of natural logarithm.
And we will get the first part as $2{{\log }_{2}}x=\dfrac{2\ln x}{\ln 2}$
And then the second part will be ${{\log }_{2}}\left( x-1 \right)=\dfrac{\ln \left( x-1 \right)}{\ln 2}$
Let us substitute them in the above obtained simplified form of the function.
As a result of this conversion, we will get the function as ${{\log }_{2}}\dfrac{{{x}^{2}}}{x-1}=\dfrac{2\ln x}{\ln 2}-\dfrac{\ln \left( x-1 \right)}{\ln 2}.$
Let us consider the terms on the right-hand side of the above equation.
In the first term, we can see that $\dfrac{2}{\ln 2}$ is the constant term that remains the same after the differentiation for it is the coefficient of the variable term.
In the second term, $\dfrac{1}{\ln 2}$ is the coefficient of the variable term as we have seen in the previous case.
Now, we will differentiate the function to find the derivative.
We will get $\dfrac{d}{dx}{{\log }_{2}}\dfrac{{{x}^{2}}}{x-1}=\dfrac{d}{dx}\left( \dfrac{2\ln x}{\ln 2}-\dfrac{\ln \left( x-1 \right)}{\ln 2} \right).$
And from this, when we use the linearity property, we will get $\dfrac{d}{dx}{{\log }_{2}}\dfrac{{{x}^{2}}}{x-1}=\dfrac{d}{dx}\dfrac{2\ln x}{\ln 2}-\dfrac{d}{dx}\dfrac{\ln \left( x-1 \right)}{\ln 2}$
Now, we will get $\dfrac{d}{dx}{{\log }_{2}}\dfrac{{{x}^{2}}}{x-1}=\dfrac{2}{\ln 2}\dfrac{d}{dx}\ln x-\dfrac{1}{\ln 2}\dfrac{d}{dx}\ln \left( x-1 \right)$
And we will get the following, $\dfrac{d}{dx}{{\log }_{2}}\dfrac{{{x}^{2}}}{x-1}=\dfrac{2}{\ln 2}\dfrac{1}{x}-\dfrac{1}{\ln 2}\dfrac{1}{x-1}.$
Hence the derivative is $\dfrac{d}{dx}{{\log }_{2}}\dfrac{{{x}^{2}}}{x-1}=\dfrac{2}{x\ln 2}-\dfrac{1}{\left( x-1 \right)\ln 2}.$

Note: We know that $\dfrac{d}{dx}\ln x=\dfrac{1}{x.}$ We also know that the linear property of differentiation is given by $\dfrac{d}{dx}\left( f\pm g \right)=\dfrac{df}{dx}\pm \dfrac{dg}{dx}$ where $f$ and $g$ are two functions of $x.$ Natural logarithm is the logarithm to the base $e,$ the constant called exponent.