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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If y = _ 10 x + _ x 10 + _ x x + _ 10 10, then (dy)/(dx) is equal to

If y=log⁡10x+log⁡x10+log⁡xx+log⁡1010,y = \log_{10} x + \log_{\mathrm{x}} 10 + \log_{\mathrm{x}} x + \log_{10} 10, then dydx\dfrac{dy}{dx} is equal to

Options

  1. A\dfrac{1}{x\log_{\mathrm{e}} 10} - \dfrac{\log_{\mathrm{e}} 10}{x{\left(\log_{\mathrm{e}} x\right)}^{2}}
  2. B1xlog⁡e10−1xlog⁡10e\dfrac{1}{x\log_{\mathrm{e}} 10} - \dfrac{1}{x\log_{10} e}
  3. C\dfrac{1}{x\log_{\mathrm{e}} 10} - \dfrac{\log_{\mathrm{e}} 10}{x\left(\log_{\mathrm{e}} x\right)}
  4. DNone of these

Correct answer

A. \dfrac{1}{x\log_{\mathrm{e}} 10} - \dfrac{\log_{\mathrm{e}} 10}{x{\left(\log_{\mathrm{e}} x\right)}^{2}}

Step-by-step solution

y = \log_{10} x + \log_{\mathrm{x}} 10 + \log_{\mathrm{x}} x + \log_{10} 10<br/><br/><br /><br />= \log_{10} x + \dfrac{\log_{\mathrm{e}} 10}{\log_{\mathrm{e}}x} + 1 + 1<br/><br/><br /><br />⇒\dfrac{dy}{dx} = \dfrac{1}{x}\log_{10} e - \dfrac{\log_{\mathrm{e}} 10}{x{\left(\log_{\mathrm{e}} x\right)}^{2}}
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