If f(x)=log _{x^{2}}left(log _{e} xright),the | vedclass

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

Question

(C) Given,$f(x) = \log _{x^{2}}(\log _{e} x) = \frac{1}{2} \log _{x}(\log _{e} x)$.Using the change of base formula,$f(x) = \frac{1}{2} \frac{\log _{e

Vedclass · Accepted Answer

$\frac{1}{2e}$

Vedclass · Answer

(C) Given,$f(x) = \log _{x^{2}}(\log _{e} x) = \frac{1}{2} \log _{x}(\log _{e} x)$.Using the change of base formula,$f(x) = \frac{1}{2} \frac{\log _{e}(\log _{e} x)}{\log _{e} x}$.Applying the quotient rule,$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\log _{e} x \cdot \frac{d}{dx}(\log _{e}(\log _{e} x)) - \log _{e}(\log _{e} x) \cdot \frac{d}{dx}(\log _{e} x)}{(\log _{e} x)^{2}} \right]$.$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\log _{e} x \cdot (\frac{1}{\log _{e} x} \cdot \frac{1}{x}) - \log _{e}(\log _{e} x) \cdot \frac{1}{x}}{(\log _{e} x)^{2}} \right]$.$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\frac{1}{x} - \frac{1}{x} \log _{e}(\log _{e} x)}{(\log _{e} x)^{2}} \right]$.At $x = e$,$\log _{e} e = 1$ and $\log _{e}(\log _{e} e) = \log _{e} 1 = 0$.Substituting these values,$f^{\prime}(e) = \frac{1}{2} \left[ \frac{\frac{1}{e} - \frac{1}{e} \cdot 0}{(1)^{2}} \right] = \frac{1}{2e}$.

If $f(x)=\log _{x^{2}}\left(\log _{e} x\right)$,then $f^{\prime}(x)$ at $x=e$ is

Similar Questions

If $f(x) = \log_{x^2}(\log_{e} x)$,then $f^{\prime}(x)$ at $x = e$ is

If $u = \log(\sqrt{x-1} - \sqrt{x+1})$ and $v = \sqrt{x+1} + \sqrt{x-1}$,then $\frac{du}{dv} = \dots$.

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

If $y=e^{\log _{e}\left[1+x+x^{2}+\ldots\right]}$,then $\frac{d y}{d x}$ is equal to

If $y=2^{ax}$ and $\left(\frac{dy}{dx}\right)_{x=1}=\log 256$,then $a=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module