If y=log _{10} x+log _x 10+log _x x+log _{10} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $y = \log_{10} x + \log_x 10 + \log_x x + \log_{10} 10$. Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we have: $y

Vedclass · Accepted Answer

$\frac{1}{x \log _e 10}-\frac{\log _e 10}{x(\log _e x)^2}$

Vedclass · Answer

(B) Given $y = \log_{10} x + \log_x 10 + \log_x x + \log_{10} 10$. Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we have: $y = \frac{\log_e x}{\log_e 10} + \frac{\log_e 10}{\log_e x} + 1 + 1$. $y = \frac{\log_e x}{\log_e 10} + \frac{\log_e 10}{\log_e x} + 2$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{\log_e 10} \cdot \frac{d}{dx}(\log_e x) + \log_e 10 \cdot \frac{d}{dx}((\log_e x)^{-1}) + 0$. $\frac{dy}{dx} = \frac{1}{\log_e 10} \cdot \frac{1}{x} + \log_e 10 \cdot (-1)(\log_e x)^{-2} \cdot \frac{1}{x}$. $\frac{dy}{dx} = \frac{1}{x \log_e 10} - \frac{\log_e 10}{x(\log_e x)^2}$.

If $y=\log _{10} x+\log _x 10+\log _x x+\log _{10} 10$,then $\frac{d y}{d x}=$

Similar Questions

If $y = \log \left(\frac{1-x^{2}}{1+x^{2}}\right)$,then $\frac{dy}{dx}$ is equal to

$\frac{d}{dx} \left[ \log \left\{ e^x \left( \frac{x - 2}{x + 2} \right)^{3/4} \right\} \right]$ is equal to

Differentiate the following with respect to $x$: $\log (\log x)$,where $x > 1$.

If $y = \log_2(\log_2 x)$,then $\frac{dy}{dx}$ is equal to

If $y = \log_2(\log_2 x)$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module