Question

Find‌ ‌the‌ ‌derivative‌ ‌of‌ ‌$y={{\log‌ ‌}_{10}}x$.‌

Answer 1

Hint: We first have to change the base of the logarithm as the derivative of logarithm \[\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\] works for base being $e$. We change the base following the rule of logarithm of ${{\log }_{a}}b=\dfrac{{{\log }_{m}}b}{{{\log }_{m}}a}$. We get $y={{\log }_{10}}x=\dfrac{\ln x}{\ln 10}$. Then we take derivatives to get \[\dfrac{1}{x\ln 10}\].

Complete step-by-step solution:
We know that differentiation of \[v\left( x \right)=\ln x\] is \[{{v}^{'}}\left( x \right)=\dfrac{1}{x}\].
We also know that $\ln a={{\log }_{e}}a$.
Therefore, we have to change the base of the logarithm from $y={{\log }_{10}}x$ to \[v\left( x \right)=\ln x\].
We know that ${{\log }_{a}}b=\dfrac{{{\log }_{m}}b}{{{\log }_{m}}a}$. We use the value of $m=e$ for $y={{\log }_{10}}x$.
$y={{\log }_{10}}x=\dfrac{{{\log }_{e}}x}{{{\log }_{e}}10}$. Now the value of ${{\log }_{e}}10$ is constant.
We can write $y={{\log }_{10}}x=\dfrac{\ln x}{\ln 10}$.
Now we find the derivative of the $y={{\log }_{10}}x=\dfrac{\ln x}{\ln 10}$ as $\dfrac{d}{dx}\left( \dfrac{\ln x}{\ln 10} \right)$ .
The constant gets out at the start. So, \[\dfrac{d}{dx}\left( \dfrac{\ln x}{\ln 10} \right)=\dfrac{1}{\ln 10}\dfrac{d}{dx}\left( \ln x \right)\].
We use the derivative formula of \[\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\] and get
\[\dfrac{d}{dx}\left( \dfrac{\ln x}{\ln 10} \right)=\dfrac{1}{\ln 10}\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x\ln 10}\].
Therefore, the derivative of $y={{\log }_{10}}x$ is \[\dfrac{1}{x\ln 10}\].

Note: If the ratio of $\dfrac{\Delta y}{\Delta x}$ tends to a definite finite limit when \[\Delta x\to 0\], then the limiting value obtained by this can also be found by first order derivative. We can also apply a first order derivative theorem to get the differentiated value of $\ln x$.
We know that $\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$. Here $f\left( x \right)=\ln x$. Also, $f\left( x+h \right)=\ln \left( x+h \right)$.
So, $\dfrac{df}{dx}=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}=\displaystyle \lim_{h \to 0}\dfrac{\ln \left( x+h \right)-\ln \left( x \right)}{h}$.
We know the limit value $\displaystyle \lim_{h \to 0}\dfrac{\ln \left( 1+\dfrac{h}{x} \right)}{\dfrac{h}{x}}\times \dfrac{1}{x}=\dfrac{1}{x}$.
Therefore, \[\dfrac{df}{dx}=\displaystyle \lim_{h \to 0}\dfrac{\ln \left( x+h \right)-\ln \left( x \right)}{h}=\dfrac{1}{x}\].