y = sec(logx)
y' = sec(logx)tan(logx)*1/secx
correct?
thank u
almost, if in doubt use the chain rule..
$\displaystyle y = \sec ( \log x)$ Let $\displaystyle u = \log x$
$\displaystyle y = \sec u \ \ \rightarrow \ \ \frac{dy}{du}= \sec u \tan u$
$\displaystyle \frac{du}{dx} = \frac{1}{x}$
$\displaystyle \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$
Bobak