# Thread: derivative help #2

1. ## derivative help #2

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

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

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

$
= \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
$
\sqrt{sec\sqrt{x}}
$

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

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

$
= \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.
$f(x)=\sqrt{sec\sqrt{x}}=(sec(x^\frac{1}{2}))^\frac {1}{2}
$

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