Question

arcsec(x/3)

Answer 1

can someone help me find the derivative?

Answer 2

Been awhile... uhhhh

Answer 3

http://www.themathpage.com/acalc/inverse-trig.htm#arcsec

Answer 4

if i get it right then..
\[
f(x) = arcsec(x) \\
g(x) = \frac{x}{3} \\
f\bigg[ g(x) \bigg]' = \bigg[ arcsec\bigg( \frac{x}{3} \bigg) \bigg]' \\
f'(x) = arcsec'(x) = \frac{1}{x \cdot \sqrt{x^2 - 1}} \\
g'(x) = 1 \cdot \frac{x^0 }{3} = \frac{1}{3} \\
f\bigg[ g(x) \bigg]' = f'\bigg[ g(x) \bigg] \cdot g'(x) = \\
= \bigg[ \frac{1}{g(x) \cdot \sqrt{g(x)^2 - 1}} \bigg] \cdot \frac{1}{3} \\
= \frac{1}{x \cdot \sqrt{ \frac{x^2}{9} - 1} }
\]