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Mathematics Question on Logarithmic Differentiation

The differential coefficient of log⁡10x\log_{10} x with respect to log⁡x10\log_{x} 10 is

A

11

B

−(log⁡10x)2-(\log_{10} x)^2

C

(log⁡x10)2(\log_{x} 10)^2

D

x2100\frac{x^2}{100}

Answer

−(log⁡10x)2-(\log_{10} x)^2

Explanation

Solution

Let u=log⁡10xu=\log _{10} x and v=log⁡x10v=\log _{x} 10
⇒u=log⁡exlog⁡e10\Rightarrow u=\frac{\log _{e} x}{\log _{e} 10} and
v=log⁡e10log⁡exv=\frac{\log _{e} 10}{\log _{e} x}
Now, dudx=1xlog⁡e10\frac{d u}{d x} =\frac{1}{x \log _{e} 10}
and dvdx=log⁡e10[−1x(log⁡ex)2]\frac{d v}{d x}=\log _{e} 10\left[\frac{-1}{x\left(\log _{e} x\right)^{2}}\right]
∴dudv=du/dxdv/dx=1xlog⁡e10÷−log⁡e10x(log⁡ex)2\therefore \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{1}{x \log _{e} 10} \div \frac{-\log _{e} 10}{x\left(\log _{e} x\right)^{2}}
=−(log⁡ex)2(log⁡e10)2=−(log⁡exlog⁡e10)2=\frac{-\left(\log _{e} x\right)^{2}}{\left(\log _{e} 10\right)^{2}}=-\left(\frac{\log _{e} x}{\log _{e} 10}\right)^{2}
=−(log⁡10x)2=-\left(\log _{10} x\right)^{2}