Question

Find these derivatives
y=arcsec(x/3)

Answer 1

@agent0smith

Answer 2

This should have everything you need: http://www.themathpage.com/acalc/inverse-trig.htm#arcsec

Answer 3

ok but i got the answer \[\frac{ 1 }{ \left| x \right| \sqrt{x^2-9} }\]

Answer 4

but the answer was \[\frac{ 3 }{ \left| x \right|\sqrt{x^2-9} }\]

Answer 5

You probably forgot to use the chain rule, with the derivative of the x/3 being 1/3.

Answer 6

but arcsecu's derivative should be \[\frac{ u' }{ \left| u \right| \sqrt{u^2-1}}\]

Answer 7

a=3
u=x
u'=1

Answer 8

so that's why i got \[\frac{ 1 }{ \left| x \right|\sqrt{x^2-9} }\] instead of \[\frac{ 3 }{ \left| x \right|\sqrt{x^2-9}}\]

Answer 9

If you're using arcsecu, then u=x/3, not u=x...

Answer 10

so then my answer would be \[\frac{ 1 }{ \left| x \right| \sqrt{\frac{ x^2 }{ 9} - 9}}\]

Answer 11

You seem to be mixing things up... you just posted \[\frac{ u' }{ \left| u \right| \sqrt{u^2-1}}\]

Answer 12

sorry top would be 1/3 instead of 1 rite?

Answer 13

u' = 1/3
u = x/3
u^2 = x^2/9

Answer 14

\[\huge \frac{ \frac{ 1 }{ 3 } }{ \left| \frac{ x }{ 3 } \right| \sqrt{\frac{ x^2 }{9 }-1}}\]

Answer 15

Now you can pull a 1/9 out of the square root...\[\huge \frac{ \frac{ 1 }{ 3 } }{ \left| \frac{ x }{ 3 } \right| \sqrt{\frac{ 1 }{ 9 }( x^2 -9)}} \]

Answer 16

so how did i end up with 3 on top?

Answer 17

the answer was \[\frac{ 3 }{ \left| x \right| \sqrt{x+1}}\]

Answer 18

Yes, take it from where I left off, simplify.

Answer 19

x^2 + 9 i mean lol

Answer 20

the answer was \[\frac{ 3 }{ \left| x \right|\sqrt{x^2-9} }\]

Answer 21

idk how to simplify to get 3 on top

Answer 22

\[\huge \frac{ \frac{ 1 }{ 3 } }{ \left| \frac{ x }{ 3 } \right| \sqrt{\frac{ 1 }{ 9 }( x^2 -9)}}\]
sqrt of 1/9 is 1/3, and dividing by a fraction is the same as multipling by it's reciprocal.

Answer 23

oh wait nvm lol

Answer 24

haha :)

Answer 25

can you help me with another one please? :)