what is the derivative of x^y = 4?

Answer 1

ln(x^y) = ln 4
y ln x = ln 4
y = ( ln 4) (ln x)^(-1)
dy/dx = (-1) (ln 4) (ln x)^(-2) (1/x)
dy/dx = (-1/x) (ln 4) (1 / ln x) ²

Answer 2

x^y = 4

take logs

yln(x) = ln(4)

differentiate

y(1/x) + ln(x)y' = 0

y'ln(x) = - y/x

y' = -y/x(1/ln(x))

substitute back y = ln(4)/ln(x)

y' = -ln(4)/ln(x)*(1/x)*(1/ln(x)) = -ln(4) / [x*(lnx)^2]

Answer 3

respect to x?

take ln for both sides
ln x^y = ln(4)

log rule:
log(a^x) = x log(a)
y ln(x) = ln(4)

y = ln(4) / ln(x)

quotient rule:
d/dx (u/v) = (u'v - v'u)/v^2
where u = ln(4) and v = ln(x)

y' = [ d/dx (ln(4)) (ln(x)) - d/dx (ln(x)) (ln(4))] / (ln(x))^2

y' = ( -1/x * ln(4)) / (ln(x))^2

y' = -ln(4)/ [x (ln(x))^2] <== answer


Rec