QUESTION

# Find the derivative of $lo{{g}_{10}}x$ with respect to ${{\log }_{x}}10$ is?

Hint: Start by letting ${{\log }_{x}}10$ as a variable. Think of a way to represent $lo{{g}_{10}}x$ in terms of ${{\log }_{x}}10$ . Once we find the relation, differentiate to reach the answer.

Before starting with the solution, let us first discuss something about the physical significance of a derivative. The derivative of a function is the measure of the steepness of the graph of the particular function at a particular point, where we are finding the derivative.
To start with the question, we let ${{\log }_{x}}10$ to be t.
We know, ${{\log }_{b}}a=\dfrac{1}{{{\log }_{a}}b}$ . Applying this formula to $lo{{g}_{10}}x$ , we get
${{\log }_{x}}10=\dfrac{1}{lo{{g}_{10}}x}=\dfrac{1}{t}$
So, now using the above results, we can say that we need to find the derivative of $\dfrac{1}{t}$ with respect to t. And we also know that the derivative $\dfrac{1}{y}$ with respect to y is $-\dfrac{1}{{{y}^{2}}}$ .
$\therefore \dfrac{d\left( \dfrac{1}{t} \right)}{dt}=\dfrac{-1}{{{t}^{2}}}$
Now we will substitute the value of t in the above equation to reach the final result. On doing so, we get
$\dfrac{d\left( \dfrac{1}{{{\log }_{x}}10} \right)}{d\left( {{\log }_{x}}10 \right)}=\dfrac{-1}{{{\left( {{\log }_{x}}10 \right)}^{2}}}$
Now as we know we can write $\left( \dfrac{1}{{{a}^{2}}} \right)\text{ as }{{\left( \dfrac{1}{a} \right)}^{2}}$ . Using the formula to the above equation, we get
$\dfrac{d\left( \dfrac{1}{{{\log }_{x}}10} \right)}{d\left( {{\log }_{x}}10 \right)}=-{{\left( \dfrac{1}{{{\log }_{x}}10} \right)}^{2}}$

Now we will again use the formula ${{\log }_{b}}a=\dfrac{1}{{{\log }_{a}}b}$ , this will give us
$\dfrac{d\left( {{\log }_{10}}x \right)}{d\left( {{\log }_{x}}10 \right)}=-{{\left( {{\log }_{10}}x \right)}^{2}}$
Therefore, we can conclude that the derivative of ${{\log }_{10}}x$ with respect to ${{\log }_{x}}10$ is $-{{\left( {{\log }_{10}}x \right)}^{2}}$ .

Note: Read the question carefully, as the general mistakes that a student makes is, as soon as they find a question related to differentiation without even reading the complete question starts to differentiate the expression with respect to x leading to errors a lot of times. Also, be careful about the signs as the mistake that a student can make while dealing with exponents include: $\left( \dfrac{-1}{{{a}^{2}}} \right)\text{ =}{{\left( \dfrac{-1}{a} \right)}^{2}}$ .