# Math Help - derivative

1. ## derivative

y = sec(logx)

y' = sec(logx)tan(logx)*1/secx

correct?

thank u

2. almost, if in doubt use the chain rule..

$y = \sec ( \log x)$ Let $u = \log x$

$y = \sec u \ \ \rightarrow \ \ \frac{dy}{du}= \sec u \tan u$

$\frac{du}{dx} = \frac{1}{x}$

$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$

Bobak

3. You have $f(x)=\sec(x), g(x)=\ln(x), and\ f\circ g(x) = \sec(\ln(x))$

So you need to use the chain rule:
$

\frac{dy}{dx}=\frac{df}{dg} \frac{dg}{dx} = \sec(g(x))\tan(g(x)) \frac{1}{x} = \frac{\sec(ln(x))\tan(ln(x))}{x}
$

4. Originally Posted by Greengoblin
You have $f(x)=\sec(x), g(x)=\ln(x), and\ f\circ g(x) = \sec(\ln(x))$

So you need to use the chain rule:
$

\frac{dy}{dx}=\frac{df}{dg} \frac{dg}{dx} = \sec(g(x))\tan(g(x)) \frac{1}{x} = \frac{\sec(ln(x))\tan(ln(x))}{x}
$

thanks for that.