Question

What is the differentiation of sec^-1 ((x^2+1)/(x^2-1))?

Answer 1

Answer on Question #50710 – Math – Differential Calculus | Equations

What is the differentiation of

$$f(x) = \sec^{-1} \left(\frac{x^2 + 1}{x^2 - 1}\right)?$$

Solution

Because (see http://en.wikipedia.org/wiki/Inverse_trigonometric_functions)

$$(\sec^{-1} t)' = \frac{1}{|t| \sqrt{t^2 - 1}}, \, |t| > 1$$

Then using the chain and quotient rules for differentiation, find

$$\begin{aligned} f'(x) = \left(\sec^{-1} \left(\frac{x^2 + 1}{x^2 - 1}\right)\right)' &= \frac{1}{\left|\frac{x^2 + 1}{x^2 - 1}\right| \sqrt{\left(\frac{x^2 + 1}{x^2 - 1}\right)^2} - 1} \cdot \left(\frac{x^2 + 1}{x^2 - 1}\right)' = \\ &= \frac{|x^2 - 1|}{|x^2 + 1|} \cdot \frac{|x^2 - 1|}{\sqrt{(x^2 + 1)^2 - (x^2 - 1)^2}} \cdot \frac{2x(x^2 - 1) - 2x(x^2 + 1)}{(x^2 - 1)^2} = \\ &= \frac{(x^2 - 1)^2}{(x^2 + 1) \sqrt{(x^2 + 1)^2 - (x^2 - 1)^2}} \cdot \frac{-4x}{(x^2 - 1)^2} = \\ &= -\frac{2x}{(x^2 + 1) |x|} = \begin{cases} -\frac{2}{x^2 + 1}, & \text{when } x > 0 \\ \frac{2}{x^2 + 1}, & \text{when } x < 0 \end{cases} \end{aligned}$$

In fact, $\left|\frac{x^2 + 1}{x^2 - 1}\right| > 1$ for all real $x, x \neq 0$.

Answer:

$$\left(\sec^{-1} \left(\frac{x^2 + 1}{x^2 - 1}\right)\right)' = -\frac{2x}{(x^2 + 1) |x|}.$$

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