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If $y=\log _{10} x+\log _{x} 10+\log _{x} x+\log _{10} 10$,
then $\frac{d y}{d x}$ is equal to
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then $\frac{d y}{d x}$ is equal to
Solution:
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Verified Answer
The correct answer is:
$\frac{1}{x \log _{e} 10}-\frac{\log _{e} 10}{x\left(\log _{e} x\right)^{2}}$
Given,
$$
y=\log _{10} x+\log _{x} 10+\log _{x} x+\log _{10} 10
$$
$\Rightarrow \quad y=\log _{10} e \cdot \log _{e} x+\frac{\log _{e} 10}{\log _{e} x}+1+1$
On differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} &=\frac{1}{x} \log _{10} e-\frac{\log _{e} 10}{x\left(\log _{e} x\right)^{2}} \\ &=\frac{1}{x \log _{e} 10}-\frac{\log _{e} 10}{x\left(\log _{e} x\right)^{2}} \end{aligned}$
$$
y=\log _{10} x+\log _{x} 10+\log _{x} x+\log _{10} 10
$$
$\Rightarrow \quad y=\log _{10} e \cdot \log _{e} x+\frac{\log _{e} 10}{\log _{e} x}+1+1$
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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