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

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

yes.

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

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

I think....

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

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

i think it would.. lets try

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

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

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

is that what you think it is?

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)

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

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

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

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

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

Then i guess you would factor out the negative?

I prefer my answer. I chain ruled it.

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

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

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

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

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

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

y is a function of x.

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