Question

Prove that d/dx arcsecu= 1/(usqrt(u^2-1)) du/dx

Answer 1

\[\frac{ d }{ dx } arcsec (u)=\frac{ 1 }{ u \sqrt{u^{2}-1} } \frac{ du }{ dx }\]

Answer 2

y=\[\sec^{-1}x\] \[secy=x\]
\[\frac{ d }{ dx } secy = \frac{ d }{ dx } x\]
\[(secytany) \frac{ du }{ dx } = 1\]

Answer 3

The du/dx on the LHS should be dy/dx, but other than that, good so far.

Answer 4

\[\frac{ dy }{ dx }= \frac{ 1 }{ secytany }\]
\[\sec^{2}y-1 = \tan^{2}y -> \sqrt{\sec^{2}y-1}=tany\]

Answer 5

\[\frac{ dy }{ dx }=\frac{ 1 }{ (secy)\sqrt{\sec^{2}y-1} } -> \frac{ 1 }{ xsqrt{x^{2}-1} } dx\]

Answer 6

Is that right????

Answer 7

oh I wrote sqrt thing in wrong lol

Answer 8

\[\frac{ 1 }{ x \sqrt{x^{2}-1} }\]

Answer 9

how do I get rid of the x

Answer 10

In your second comment, you let y = arcsec(x). In fact, it should be y = arcsec(u). Just make the switch and it all works out.