How do you differentiate f(x)=arccos(tan(1/(1-x^2)) ) using the chain rule?

Jan 29, 2016

$f ' \left(x\right) = - \frac{2 x {\sec}^{2} \left(\frac{1}{1 - {x}^{2}}\right)}{{\left(1 - {x}^{2}\right)}^{2} \sqrt{1 - {\tan}^{2} \left(\frac{1}{1 - {x}^{2}}\right)}}$

Explanation:

To differentiate $\arccos$ functions, we can use the chain rule to state that

$\frac{d}{\mathrm{dx}} \arccos \left(u\right) = - \frac{1}{\sqrt{1 - {u}^{2}}} \cdot u '$

Here, $u = \tan \left(\frac{1}{1 - {x}^{2}}\right)$, so

$f ' \left(x\right) = - \frac{1}{\sqrt{1 - {\tan}^{2} \left(\frac{1}{1 - {x}^{2}}\right)}} \cdot \frac{d}{\mathrm{dx}} \tan \left(\frac{1}{1 - {x}^{2}}\right)$

Now, to differentiate the $\tan$ function, we use the chain rule again:

$\frac{d}{\mathrm{dx}} \tan \left(u\right) = {\sec}^{2} \left(u\right) \cdot u '$

This time, $u = \frac{1}{1 - {x}^{2}} = {\left(1 - {x}^{2}\right)}^{-} 1$, giving

$f ' \left(x\right) = - \frac{1}{\sqrt{1 - {\tan}^{2} \left(\frac{1}{1 - {x}^{2}}\right)}} \cdot {\sec}^{2} \left(\frac{1}{1 - {x}^{2}}\right) \frac{d}{\mathrm{dx}} {\left(1 - {x}^{2}\right)}^{-} 1$

Before we find the next derivative, we can simplify a little.

$f ' \left(x\right) = - {\sec}^{2} \frac{\frac{1}{1 - {x}^{2}}}{\sqrt{1 - {\tan}^{2} \left(\frac{1}{1 - {x}^{2}}\right)}} \cdot \frac{d}{\mathrm{dx}} {\left(1 - {x}^{2}\right)}^{-} 1$

To find this final derivative, use the chain rule once more:

$\frac{d}{\mathrm{dx}} {u}^{-} 1 = - {u}^{-} 2 \cdot u '$

This time around, $u = 1 - {x}^{2}$. Thus,

$f ' \left(x\right) = - {\sec}^{2} \frac{\frac{1}{1 - {x}^{2}}}{\sqrt{1 - {\tan}^{2} \left(\frac{1}{1 - {x}^{2}}\right)}} \cdot - {\left(1 - {x}^{2}\right)}^{-} 2 \frac{d}{\mathrm{dx}} \left(1 - {x}^{2}\right)$

Since $\frac{d}{\mathrm{dx}} \left(1 - {x}^{2}\right) = - 2 x$, this whole mess simplifies to be

$f ' \left(x\right) = - \frac{2 x {\sec}^{2} \left(\frac{1}{1 - {x}^{2}}\right)}{{\left(1 - {x}^{2}\right)}^{2} \sqrt{1 - {\tan}^{2} \left(\frac{1}{1 - {x}^{2}}\right)}}$