# What is the derivative of tan^-1(7x^2-1)^(1/2)?

May 16, 2015

Chain rule time here!

But first let's just remember that ${\tan}^{-} 1 \left(x\right) = \arctan \left(x\right)$.

Rewriting your expression, we get $y = \arctan {\left(7 {x}^{2} - 1\right)}^{\frac{1}{2}}$

We can name $v = 7 {x}^{2} - 1$ and $u = {v}^{\frac{1}{2}}$. Thus, $y = \arctan \left(u\right)$.

So, now, by chain rule definition,

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

Going part by part, here:

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{u '}{1 + {u}^{2}}$
$\frac{\mathrm{du}}{\mathrm{dv}} = \left(\frac{1}{2}\right) \cdot \left(\frac{1}{v} ^ \left(\frac{1}{2}\right)\right)$
$\frac{\mathrm{dv}}{\mathrm{dx}} = 14 x$

Now, getting back to the chain rule definition, we have this:

$\frac{u '}{1 + {u}^{2}} \left(\frac{1}{\cancel{2}}\right) \left(\frac{1}{v} ^ \left(\frac{1}{2}\right)\right) \cancel{14} 7 x$

After cancelling the $2$ and simplifying $14$ as $7$, we can now substitute the values of $u$ and $v$, as our objective is to get the answer as function of $x$.

$\frac{\left(\frac{1}{2}\right) \cdot \left(\frac{1}{v} ^ \left(\frac{1}{2}\right)\right)}{1 + {\left({v}^{\frac{1}{2}}\right)}^{2}} \cdot \frac{1}{7 {x}^{2} - 1} \cdot 7 x$
$\frac{\frac{1}{2} {\left(7 {x}^{2} - 1\right)}^{\frac{1}{2}}}{\cancel{1} + 7 {x}^{2} \cancel{- 1}} \cdot \frac{1}{7 {x}^{2} - 1} \cdot 7 x$

We can proceed to some more cancellings:

$\frac{\frac{1}{2} {\left(7 {x}^{2} - 1\right)}^{\frac{1}{2}}}{7 {x}^{2}} \cdot \frac{1}{7 {x}^{\cancel{2}} - 1} \cdot \cancel{7 x}$

And, now, rearrange:

((7x^2-1)^(1/2)/2)/(7x^2(7x-1))=color(green)((7x^2-1)^(1/2))/color(green)(14x^2(7x-1)