Question

# The differential coefficient of ${\log _{10}}x$ with respect to ${\log _x}10$ is A. 1B. $- {({\log _{10}}x)^2}$ C. ${({\log _x}10)^2}$ D. $\dfrac{{{x^2}}}{{100}}$

Hint: Differential coefficient is nothing but finding out the derivative of a function with respect to the function which is given.

We have to find out the differential coefficient of ${\log _{10}}x$ with respect to ${\log _x}10$

Let us consider y=${\log _{10}}x$ and z= ${\log _x}10$

So, as per the question ,we have to find out the derivative of y with respect to z

So, we have to find $\dfrac{{dy}}{{dz}}$
Since, we cannot find out the value of $\dfrac{{dy}}{{dz}}$directly letâ€™s multiply y and z

So, we get yz=(${\log _{10}}x$)(${\log _x}10$)

Now, letâ€™s make use of the formula ${\log _b}a = \dfrac{{\log a}}{{\log b}}$ and express yz in this form

So, we get $yz = \dfrac{{\log x}}{{\log 10}} \times \dfrac{{\log 10}}{{\log x}} = 1$

So, we have got yz=1

From this, we get $y = \dfrac{1}{z}$

Now, let us differentiate y with respect to z

So, we get $\dfrac{{dy}}{{dz}} = - \dfrac{1}{{{z^2}}}\left( {\because \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = - \dfrac{1}{{{x^2}}}} \right)$

The value of $\dfrac{1}{z} = y$

So, we get $\dfrac{{dy}}{{dz}}$=$- {y^2} = - {({\log _{10}}x)^2}$

Since $\dfrac{1}{z} = y$

So, option B is the correct answer for this question

Note: Make use of the appropriate formula of logarithms wherever needed and solve the question and also give importance to the function with respect to which the given has to be differentiated.