Question

find dy/dx if y= (cos x) ^ x

Answer 1

ln is your friend

Answer 2

or is it e?

Answer 3

y= (cos x) ^ x
it is easier if there is no index
so take ln

ln y = ln (cosx)^x
lny=xlncosx
diff. both sides w.r.t. x
(1/y)(dy/dx) = lncosx+ x(1/cosx)(-sinx)
=ln cosx - tanx
dy/dx=(ln cosx - tanx)(y)
=(ln cosx - tanx)( (cos x) ^ x)

Answer 4

amendment:
(1/y)(dy/dx) = lncosx+ x(1/cosx)(-sinx)
=ln cosx - xtanx
dy/dx=(ln cosx -x tanx)(y)
=(ln cosx - xtanx)( (cos x) ^ x)

Answer 5

e^y = e^(x cos(x))

e^y y'= (x cos(x))' e^(x cos(x))

e^y y'= (cos(x)-xsin(x)) e^(x cos(x))

e^y y'= (cos(x)-xsin(x)) e^y

y'= cos(x)-xsin(x)

hmmm, that is soo wrong

Answer 6

\[y'=-xsin(x)\ln(cosx)\]

Answer 7

ln(y) = x ln(cos(x))

y'/y = ln(cos(x)) - x tan(x)

y' = (cos(x))^x (ln(cos(x)) - x tan(x))