Question

The derivative of ${\sec }^{-1}\left(\frac{1}{2x^2+1}\right)\mathrm{w}.\mathrm{r}.\mathrm{t}.\sqrt{1+3x}\mathrm{at}x=-1/3$

(a) does not exist
(b) 0
(c) 1/2
(d) 1/3

Answer 1

(a) does not exist

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}{\sec }^{-1}\alpha \mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}\alpha \in \left(-1,1\right) \\ \mathrm{Here}\mathrm{for}x=\frac{-1}{3},\frac{1}{2x^2+1}=\frac{9}{11}\in \left(-1,1\right) \\ \therefore {\sec }^{-1}\left(\frac{1}{2x^2+1}\right)\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{at}x=\frac{-1}{3} \\ \therefore \mathrm{Derivative}\mathrm{of}{\sec }^{-1}\left(\frac{1}{2x^2+1}\right)\mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{at}x=\frac{-1}{3} \end{gathered}$$