Question

Derivative of ${\log _{10}}x$ with respect to ${x^2}$ is;
$\left( 1 \right)2{x^2}{\log _e}10$
$\left( 2 \right){\log _{10}}\dfrac{e}{{2{x^2}}}$
$\left( 3 \right){\log _e}\dfrac{{10}}{{2{x^2}}}$
$\left( 4 \right){x^2}{\log _e}10$

Answer 1

This question demands the knowledge of the concept of differentiation of one function with respect to another function. Let the two functions are $f\left( x \right){\text{ and g}}\left( x \right)$ respectively and according to the concept we have to calculate the value of $\dfrac{{df\left( x \right)}}{{dg\left( x \right)}}$ , which can also be written as $\left[ {\dfrac{{\dfrac{{df\left( x \right)}}{{dx}}}}{{\dfrac{{dg\left( x \right)}}{{dx}}}}} \right]$ means ultimately we have to calculate $\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ .

Complete step by step answer:
Since the two given functions have to be differentiated individually, therefore;
Let $u = {\log _{10}}x$ and $v = {x^2}$
Let us calculate the value of $\dfrac{{du}}{{dx}}{\text{ and }}\dfrac{{dv}}{{dx}}$;
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\log }_{10}}x} \right){\text{ }}......\left( 1 \right)$
By change of base formula of logarithm, we know that;
$\Rightarrow {\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$
Applying this rule in equation $\left( 1 \right)$ , we get;
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{{{\log }_e}x}}{{{{\log }_e}10}}} \right)$
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{{{\log }_e}10}} \times \left( {\dfrac{1}{x}} \right)$
We get the value of $\dfrac{{du}}{{dx}}$ , as;
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{{{\log }_e}10}}\left( {\dfrac{1}{x}} \right){\text{ }}......\left( 2 \right)$
Similarly, we will find the value of $\dfrac{{dv}}{{dx}}$;
$\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2}} \right)$
By the differentiation formula , i.e. ${x^n} = n{x^{n - 1}}$ , we get;
$\Rightarrow \dfrac{{dv}}{{dx}} = 2x{\text{ }}......\left( 3 \right)$
From equation $\left( 2 \right)$ and equation $\left( 3 \right)$, we have the values of $\dfrac{{du}}{{dx}}{\text{ and }}\dfrac{{dv}}{{dx}}$;
According to the given question we were asked to differentiate ${\log _{10}}x$ with respect to ${x^2}$ is;
Therefore, dividing equation ${\text{equation}}\left( 2 \right){\text{ by equation }}\left( 3 \right){\text{ , we get;}}$
$\Rightarrow \dfrac{{\dfrac{{du}}{{dx}}}}{{\dfrac{{dv}}{{dx}}}} = \dfrac{{\dfrac{1}{{{{\log }_e}10}}\left( {\dfrac{1}{x}} \right)}}{{2x}}$
On further simplification of the above equation, we get;
$\Rightarrow \dfrac{{du}}{{dv}} = \dfrac{1}{{{{\log }_e}10}}\left( {\dfrac{1}{x} \times \dfrac{1}{{2x}}} \right)$
$\Rightarrow \dfrac{{du}}{{dv}} = \dfrac{1}{{{{\log }_e}10}}\left( {\dfrac{1}{x} \times \dfrac{1}{{2x}}} \right)$
Therefore the value of $\dfrac{{du}}{{dv}}$ equals to;
$\Rightarrow \dfrac{{du}}{{dv}} = \dfrac{1}{{{{\log }_e}10}}\left( {\dfrac{1}{{2{x^2}}}} \right){\text{ }}......\left( 4 \right)$
But according to the options given to us , we will have to further simplify the above equation;
By the logarithmic property, we know that;
$\Rightarrow {\log _a}b = \dfrac{1}{{{{\log }_b}a}}$
Applying the above property in equation $\left( 4 \right)$, we get;
$\Rightarrow \dfrac{{du}}{{dv}} = \dfrac{1}{{\dfrac{1}{{{{\log }_e}10}} \times 2{x^2}}}$
Rearranging the above equation, we get;
$\Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{{{\log }_{10}}e}}{{2{x^2}}}$
Therefore, the differentiation of ${\log _{10}}x$ with respect to ${x^2}$ is; $\dfrac{{du}}{{dv}} = \dfrac{{{{\log }_{10}}e}}{{2{x^2}}}$

So, the correct answer is “Option 2”.

Note: The change of base formula of logarithm becomes very important in cases where we have to calculate the answer according to the given options such as in this case. Let see an example for more clarity on this concept; suppose we have to calculate the value of ${\log _3}8$ , we can not calculate it directly using a calculator that is where the change of base formula comes into the picture and it is given as ; $\Rightarrow {\log _a}b = \dfrac{{{{\log }_x}b}}{{{{\log }_x}a}}$ , upon conversion $x{\text{ can represent any base}}$, but it must be same for both numerator and denominator. Coming back to our example: ${\log _3}8 = \dfrac{{{{\log }_e}8}}{{{{\log }_e}3}}$ , which gives 1.8928 approximately. No matter what base we take we will always get the same answer.