1. ## derivative help #2

$\displaystyle \sqrt{sec\sqrt{x}}$

$\displaystyle = (sec(x^\frac{1}{2}))^\frac{1}{2}$

$\displaystyle = \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2}$

$\displaystyle = \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2} \cdot sec(x)^\frac{1}{2}tan(x)^\frac{1}{2} \cdot \frac{1}{2}x^\frac{-1}{2}$

this the right path? thank u

2. Originally Posted by jvignacio
$\displaystyle \sqrt{sec\sqrt{x}}$

$\displaystyle = (sec(x^\frac{1}{2}))^\frac{1}{2}$

$\displaystyle = \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2}$

$\displaystyle = \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2} \cdot sec(x)^\frac{1}{2}tan(x)^\frac{1}{2} \cdot \frac{1}{2}x^\frac{-1}{2}$

this the right path? thank u
I think your answer is correct. It's just the way you present it. It doesn't make sense to put all those equal signs because they are clearly are not all equal.
$\displaystyle f(x)=\sqrt{sec\sqrt{x}}=(sec(x^\frac{1}{2}))^\frac {1}{2}$

Then $\displaystyle f'(x)=\frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2} \times (sec(x^\frac{1}{2}))'$
$\displaystyle = \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2} \cdot sec(x)^\frac{1}{2}tan(x)^\frac{1}{2} \cdot \frac{1}{2}x^\frac{-1}{2}[$