MHT CET · Maths · Differentiation
If \(y=\log _{10} x+\log _{x} 10+\log _{x} x+\log _{10} 10\),
then \(\frac{d y}{d x}\) is equal to
- A \(\frac{1}{x \log _{e} 10}-\frac{\log _{e} 10}{x\left(\log _{e} x\right)^{2}}\)
- B \(\frac{1}{x \log _{e} 10}-\frac{1}{x \log _{10} e}\)
- C \(\frac{1}{x \log _{e} 10}+\frac{\log _{e} 10}{x\left(\log _{e} x\right)^{2}}\)
- D None of the above
Answer & Solution
Correct Answer
(A) \(\frac{1}{x \log _{e} 10}-\frac{\log _{e} 10}{x\left(\log _{e} x\right)^{2}}\)
Step-by-step Solution
Detailed explanation
Given,
\(
y=\log _{10} x+\log _{x} 10+\log _{x} x+\log _{10} 10
\)
\(\Rightarrow y=\log _{10} e \cdot \log _{e} x+\frac{\log _{e} 10}{\log _{e} x}+1+1\)
On differentiating w.r.t. \(x\), we get
\(\begin{aligned} \frac{d y}{d x} &=\frac{1}{x} \log _{10} e-\frac{\log _{e} 10}{x\left(\log _{e} x\right)^{2}} \\ &=\frac{1}{x \log _{e} 10}-\frac{\log _{e} 10}{x\left(\log _{e} x\right)^{2}} \end{aligned}\)
\(
y=\log _{10} x+\log _{x} 10+\log _{x} x+\log _{10} 10
\)
\(\Rightarrow y=\log _{10} e \cdot \log _{e} x+\frac{\log _{e} 10}{\log _{e} x}+1+1\)
On differentiating w.r.t. \(x\), we get
\(\begin{aligned} \frac{d y}{d x} &=\frac{1}{x} \log _{10} e-\frac{\log _{e} 10}{x\left(\log _{e} x\right)^{2}} \\ &=\frac{1}{x \log _{e} 10}-\frac{\log _{e} 10}{x\left(\log _{e} x\right)^{2}} \end{aligned}\)
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