Question
If $\text{f(x)}=\log_{\text{x}^2}(\log\text{x}),$ the f(x) at x = e is:
- $0$
- $1$
- $\frac{1}{\text{e}}$
- $\frac{1}{2\text{e}}$
Solution:
We have, $\text{f(x)}=\log_{\text{x}^2}(\log\text{x})$
$\Rightarrow\text{f(x)}=\frac{\log(\log\text{x})}{\log\text{x}^2}$
$\Rightarrow\text{f(x)}=\frac{\log(\log\text{x})}{2\log\text{x}}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\frac{\text{d}}{\text{dx}}\bigg\{\frac{\log(\log\text{x})}{\log\text{x}}\bigg\}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\log\text{x}}\times\frac{1}{\text{x}}\times\log\text{x}-\frac{\log(\log)\text{x}}{\text{x}}}{(\log\text{x})^2}\Bigg\}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\text{x}}-\frac{\log(\log)\text{x}}{\text{x}}}{(\log\text{x})^2}\Bigg\}$
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\text{e}}-\frac{\log(\log)\text{e}}{\text{e}}}{(\log\text{e})^2}\Bigg\}$
[Putting x = e]
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{2}\times\bigg\{\frac{\frac{1}{\text{e}}}{1}\bigg\}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2\text{e}}$
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