Question

Find the derivative

Answer 1

g(x) = (secx)^x

Answer 2

\[\log g(x)=x \log \sec x\]
find derivative

Answer 3

\[\frac{ g \prime(x) }{ g(x) }=?(by~product~rule)\]

Answer 4

u=x
,v=log sec x

Answer 5

\[\frac{ d }{ dx }\left( uv \right)=u v \prime+u \prime v\]

Answer 6

\[\frac{ d }{ dx }(\sec x)=\frac{ \sec x \tan x }{ \sec x }\]

Answer 7

We've been using ln in class, could I use that interchangeably with log? I'm a little confused on the second step... where do you get the g'(x)/g(x) from? Sorry, I'm new at this :P

Answer 8

correction\[\frac{ d }{ dx }\left( \log \sec x \right)=\frac{ \sec x \tan x }{ \sec x }\]

Answer 9

\[\frac{ d }{ dx }\left( \log g(x) \right)=\frac{ g \prime(x) }{ g(x) }\]

Answer 10

\[\frac{ d }{ dx }\left\{ \log x \right\}=\frac{ 1 }{ x }\]
log x is natural log

Answer 11

I'm still confused on this :/ can you start from the beginning and explain a bit more? I really appreciate the help

Answer 12

\[\large y = (\sec x)^x\]take ln of both sides\[\large \ln y = \ln (\sec x)^x\]simplify\[\large \ln y = x\ln (\sec x)\]Now differentiate.

Answer 13

\[\frac{ 1 }{ y }*y' = x*1/secx + \ln(secx)\]

Answer 14

Not sure if I did that right. Thanks angel!

Answer 15

\[\large \ln y = x\ln (\sec x) \]Remember deriv of \(\large \ln f(x)\) is \(\Large \frac{ f'(x) }{ f(x) }\) and use product rule on right... i left the deriv of the \(\large \ln (\sec x)\) to the next step
\[\large \frac{ y' }{ y } = 1*\ln (\sec x) +x[\ln (\sec x)]'\]
\[\large \frac{ y' }{ y } = 1*\ln (\sec x) +x\frac{ \sec x \tan x }{ \sec x }\]

Answer 16

Ahhh I see! I was taking the derivative of ln(secx) wrong. Thank you so much Age <3
Next I would just multiply that y to the right side?

Answer 17

would it help to know that \[\sec(x)=\frac{1}{\cos(x)}\] making \[\log(\sec(x))=-\log(\cos(x)\]?

Answer 18

Yep. After simplifying the above:\[\large y' =y (\ln (\sec x) +x \tan x) \]remember\[\large y = (\sec x)^x \]so\[\large y' =(\sec x)^x [\ln (\sec x) +x \tan x] \]

Answer 19

Perfect! Thank you so much. It makes sense now.

Answer 20

What @satellite73 gave could help a little too, makes it slightly simpler.

Answer 21

Satellite, that does help! Thank you

Answer 22

What property is that? I don't think I've ever seen it before..

Answer 23

\[\large \ln a^b = b \ln a\]It's just the exponent property of logs\[\large \ln \frac{ 1 }{ a } = \ln a^{-1} = -1 \ln a\]

Answer 24

\[\large \ln (\sec x) = \ln \frac{ 1 }{ \cos x } = \ln (\cos x)^{-1} = -1 \ln (\cos x)\]

Answer 25

Aha :D cool. thanks for explaining!

Answer 26

:)