KCET · Maths · Differentiation
The differential coefficient of \( \log _{10} x \) with respect to \( \log _{x} 10 \) is
- A \( 1 \)
- B \( -\left(\log _{10} x\right)^{2} \)
- C \( \left(\log _{x} 10\right)^{2} \)
- D \( \frac{x^{2}}{100} \)
Answer & Solution
Correct Answer
(B) \( -\left(\log _{10} x\right)^{2} \)
Step-by-step Solution
Detailed explanation
Let \( y=\log _{10} x \) and \( z=\log _{x} 10 \)
We know that, \( \log _{a} b=\frac{\log b}{\log a} \)
So, \( y=\frac{\log x}{\log 10} \) and \( z=\frac{\log 10}{\log x} \)
Thus, \( y=\frac{1}{z} \)
\( \Rightarrow \frac{d y}{d z}=\frac{-1}{z^{2}}=-\left(\frac{1}{z}\right)^{2}=-y^{2} \)
\( \Rightarrow \frac{d y}{d x}=-\left(\log _{10} x\right)^{2} \)
We know that, \( \log _{a} b=\frac{\log b}{\log a} \)
So, \( y=\frac{\log x}{\log 10} \) and \( z=\frac{\log 10}{\log x} \)
Thus, \( y=\frac{1}{z} \)
\( \Rightarrow \frac{d y}{d z}=\frac{-1}{z^{2}}=-\left(\frac{1}{z}\right)^{2}=-y^{2} \)
\( \Rightarrow \frac{d y}{d x}=-\left(\log _{10} x\right)^{2} \)
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