Question

If $ y=log_{10}\,x+log_{x}\,10+log_{x}\,x+log_{10}\,10 $ , then $ \frac{dy}{dx} $ 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 1

Answer: (A)

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
$\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}} $