Question

find the derivative of F(x)=(ln(3))^(cos(x))

Answer 1

\[\ln(3)^{\cos(x)}\]

Answer 2

yes.

Answer 3

Write the equation like \[y = \ln(3)^{\cos(x)}\]

Answer 4

-log(3)^(cos(x))*log(log(3))*sin(x)

Answer 5

I think....

Answer 6

i was thinking taking natural log on both sides but i dont think that would help??

Answer 7

yeah it won't because i'm not solving for x, i'm just trying to find the derivative.

Answer 8

i think it would.. lets try

Answer 9

\[\ln(y) = \cos(x) \ln(\ln(3))\]

Answer 10

My answer is correct. The chain rule I used is a little overly complicated though.

Answer 11

\[-\log(3)^{cosx}*(\log(\log(3))) *\sin(x)\]

Answer 12

is that what you think it is?

Answer 13

u = cos(x); F'(x) = d(log^u(3))/du * du/dx; du/dx = -sin(x); d(log(3)^u)/du = log(3)^u * log(log(3)); F'(x) = log(3)^u * log(log(3)) * -sin(x)

Answer 14

\[\ln (\ln 3) -\sin (x) \ln^{cos(x)}(3)\]

Answer 15

\[ln(y) = -\ln(\ln(3))\sin(x)\]

Answer 16

\[\frac{y \prime}{y} = -\ln(\ln(3))\sin(x)\]

Answer 17

\[y \prime = y[-\ln(\ln(3))\sin(x)]\]

Answer 18

\[y \prime = \ln(3)^{\cos(x)}[-\ln(\ln(3))\sin(x)]\]

Answer 19

Then i guess you would factor out the negative?

Answer 20

I prefer my answer. I chain ruled it.

Answer 21

i'm confused on how you went from ln(y) to y'/y

Answer 22

the derivative of \[\ln(y)\] is \[\frac{1}{y} * y \prime\]

Answer 23

i thought the derivative of ln(x)=(1/x)?

Answer 24

It's the chain rule of that function. the derivative of ln(x) = 1/x, so using the chain rule with u = y, d(ln(u))/du * du/dx = 1/u * y' = y'/u = y'/y

Answer 25

you have to use implicit differentiation because ln is a function of y

Answer 26

or y is a function of ln.. how ever it goes

Answer 27

y is a function of x.