Consider the following statements:Statement 1 | vedclass

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

Question

(A) For Statement $1$:$y = \log_{10} x + \log_{e} x$Using the change of base formula,$\log_{a} b = \frac{\log_{e} b}{\log_{e} a}$,we have:$y = \frac{\

Vedclass · Accepted Answer

Statement $1$ is true,Statement $2$ is false.

Vedclass · Answer

(A) For Statement $1$:$y = \log_{10} x + \log_{e} x$Using the change of base formula,$\log_{a} b = \frac{\log_{e} b}{\log_{e} a}$,we have:$y = \frac{\log_{e} x}{\log_{e} 10} + \log_{e} x$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{x \log_{e} 10} + \frac{1}{x}$Since $\frac{1}{\log_{e} 10} = \log_{10} e$,we get:$\frac{dy}{dx} = \frac{\log_{10} e}{x} + \frac{1}{x}$.Thus,Statement $1$ is true.For Statement $2$:$\frac{d}{dx}(\log_{10} x) = \frac{d}{dx} \left( \frac{\log_{e} x}{\log_{e} 10} \right) = \frac{1}{x \log_{e} 10}$.The statement claims it is $\frac{\log x}{\log 10}$,which is incorrect.Similarly,$\frac{d}{dx}(\log_{e} x) = \frac{1}{x}$.The statement claims it is $\frac{\log x}{\log e}$,which is incorrect.Thus,Statement $2$ is false.

Consider the following statements:
Statement $1$: If $y = \log_{10} x + \log_{e} x$,then $\frac{dy}{dx} = \frac{\log_{10} e}{x} + \frac{1}{x}$.
Statement $2$: $\frac{d}{dx}(\log_{10} x) = \frac{\log x}{\log 10}$ and $\frac{d}{dx}(\log_{e} x) = \frac{\log x}{\log e}$.

Similar Questions

$\frac{d}{dx} \left( \log \sqrt{\frac{1+\sin x}{1-\sin x}} \right) = $

Find the derivative: $\frac{d}{dx} \log (\sqrt{x - a} + \sqrt{x - b})$

For $x^2-4 \neq 0$,the value of $\frac{d}{d x}\left[\log \left\{e^x\left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}\right\}\right]$ at $x=3$ is

If $y = \log \log x$,then $e^y \frac{dy}{dx} = $

$y = \log \left( \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x} \right) \Rightarrow \frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Consider the following statements:Statement $1$: If $y = \log_{10} x + \log_{e} x$,then $\frac{dy}{dx} = \frac{\log_{10} e}{x} + \frac{1}{x}$.Statement $2$: $\frac{d}{dx}(\log_{10} x) = \frac{\log x}{\log 10}$ and $\frac{d}{dx}(\log_{e} x) = \frac{\log x}{\log e}$.