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

$\frac{d}{dx}(\log \tan x) = $

Differentiate the following with respect to $x$: $\log(\cos e^{x})$

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

Mrs. Rajni deposited $Rs. 10,000$ in a bank that pays $4 \%$ interest compounded continuously. The amount she gets after $10$ years is approximately $Rs.$ . . . . . . . (Given $e^{(0.4)} = 1.49182$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module