use logrithmic differentiation to find the derivative of y=x^ square root of x?

Answer 1

y = x^sqrt(x)
ln y = (x^0.5) ln x
Differentiate implicitly:
(1/y)y' = (0.5x^-0.5) ln x + (x^0.5)(1/x)
(1/y)y' = (0.5x^-0.5) ln x + x^-0.5
(1/y)y' = (x^-0.5)(0.5 ln x + 1)
y' = y(x^-0.5)(0.5 ln x + 1)
y' = (x^sqrt(x))(x^-0.5)(0.5 ln x + 1)
Or y' = x^(sqrt(x)-0.5) * (0.5 ln x + 1).

Answer 2

Sorry, yes I did a mistake =(
Such a baby mistake jeje... Regards everybody!

y= x^√x

We apply ln in each side:
ln y = √x lnx

And derivate implicity:

y' /y = √x /x + 1/2 x^(-1/2) lnx


y'/y = x^(-1/2) + 1/2 x^(-1/2) lnx

We factorize and sustitute y = x^√x

y' = x^(-1/2) ( 1 + 1/2 lnx) (x^√x)

y'= 1 + 1/2 lnx

Answer 3

Stephen M is correct, but an easier way to do it would be to put the equation to base e and work from there. The equation in base e form would be y=e^((x^.5)(ln(x)).

Answer 4

y = x^(sqrt(x)). Let u=logy. Then

u = log y = sqrt(x)*logx
du/dx = 0.5logx/sqrt(x) + sqrt(x)/x

and y=e^u so dy/du = e^u = y = x^(sqrt(x)).
So dy/dx = dy/du * du/dx = x^(sqrt(x)) * (0.5logx/sqrt(x) + 1/sqrt(x))
= x^(sqrt(x)) * (0.5logx + 1)/sqrt(x)
.

I think.