Test Index

AP EAPCET 23-Aug-2021 Shift 2 Paper

Section: Mathematics

© examsnet.com

Question : 68 of 160

Marks: +1, -0

If

f(x)=\log_{x^{2}}(\log x)

f'(x)

x=e

is _____

0
1
$e^{-1}$
$(2e)^{-1}$

Solution:

f(x)=\log_{x^{2}}(\log x)

\therefore f'(x)=\frac{d}{dx}\log_{x^{2}}(\log x)

=\frac{d}{dx}\left\{\frac{\log(\log x)}{\log(x^{2})}\right\}

[\because \frac{d}{dx}\frac{u}{v}=\frac{vu'-uv'}{v^{2}}, \frac{d}{dx}\log x=\frac{1}{x}

and

\log_{a}b=\frac{\log_{e}b}{\log_{e}a}]

=\frac{1}{2}\left[\frac{\log x \cdot \frac{1}{\log x} \cdot \frac{1}{x} - \log(\log x) \cdot \frac{1}{x}}{(\log x)^{2}}\right\}

=\frac{1}{2}\left[\frac{1}{x}\frac{1-\log(\log x)}{(\log x)^{2}}\right]

At,

x=e

f'(e)=\frac{1}{2e(\log_{c}e)^{2}}(1-\log_{e}(\log_{e}e))

f'(e)=\frac{1}{2e\cdot (1)^{2}}(1-0)=\frac{1}{2e}=(2e)^{-1}

© examsnet.com

Go to Question:

Prev Question

Next Question