Practicing Success
If f(x) = logx(logx), then f'(x) at x = e is equal to: |
e 1/e 2/e 0 |
1/e |
$y=\log _x(\log x)=\frac{\log (\log x)}{\log x}=\frac{\log t}{t}$, where t = log x ∴ when x = e, t = 1 $\frac{d y}{d t}=\frac{t . \frac{1}{t}-(\log t)}{t^2}-1=\frac{1-\log t}{t^2}$ $\Rightarrow \frac{d y}{d x}=\frac{d y}{d t} \cdot \frac{d t}{d x}=\frac{1-\log t}{t^2} \times \frac{1}{x}$ $\left.\frac{d y}{d x}\right|_{x=e}=\frac{1-\log 1}{(1)^2} \times \frac{1}{e}=\frac{1}{e}$ Hence (2) is correct answer. |