# How do you take the derivative of sec^-1 (3x^2)?

Aug 26, 2015

${y}^{'} = \frac{2}{x \cdot \sqrt{9 {x}^{4} - 1}}$

#### Explanation:

For your function $y = {\sec}^{- 1} \left(3 {x}^{2}\right)$, you know that you can write

$\sec \left(y\right) = 3 {x}^{2}$

Use implicit differentiation and the fact that

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(\sec x\right) = \sec x \cdot \tan x}$

to write

$\frac{d}{\mathrm{dy}} \left(\sec y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(3 {x}^{2}\right)$

$\sec y \cdot \tan y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 6 x$

Isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$ on one side of the equation to get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x}{\sec y \cdot \tan y}$

Now, $\text{arcsec} \left(x\right)$ has a range of $\left[0 , \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2} , \pi\right]$. If you evaluate the product of $\sec x$ and $\tan x$ over the first interval, $\left[0 , \frac{\pi}{2}\right)$, you will always end up with a positive number because both the value of the tangent, and the value of secant are positive in that interval.

For the second interval, $\left(\frac{\pi}{2} , 0\right]$, the product will once again be positive, since the tangent and the secant are both negative in that interval.

The $\sec x \cdot \tan x$ product will always be positive, which means that you can write

$\sec x \cdot \tan x = \sqrt{{\left(\sec x \cdot \tan x\right)}^{2}}$

Take this back to your derivative calculation

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x}{\sqrt{{\left(\sec y \cdot \tan y\right)}^{2}}} = \frac{6 x}{\sqrt{{\sec}^{2} y \cdot {\tan}^{2} y}}$

Another important thing to remember is that

color(blue)(sec^2x = 1 + tan^2x

which means that you have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x}{\sqrt{{\sec}^{2} y \cdot \left({\sec}^{2} y - 1\right)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x}{| {\sec}^{2} y | \cdot \sqrt{{\sec}^{2} y - 1}}$

Since $\sec y = 3 {x}^{2}$, you get that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x}{{\underbrace{3 {x}^{2}}}_{\textcolor{red}{\text{always positive}}} \cdot \sqrt{{\left(3 {x}^{2}\right)}^{2} - 1}}$

Finally, the derivative of $y = {\sec}^{- 1} \left(3 {x}^{2}\right)$ is equal to

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{g r e e n}{\frac{2}{x \cdot \sqrt{9 {x}^{4} - 1}}}$