Problem: y=sqrtx ^1/x. Find DY/DX.

ln(f(x))=lnx^1/2x or lnx to the 1/2x

1/2x lnx

product rule

(1/4x^2)(lnx)+(1/x)(1/2x)
is the product rule correct?
What do i do after this product rule?

y = √(x^(1/x)) = x^(1/2x)

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

Yes, the product rule is correct. After applying the product rule to the function ln(f(x)) = (1/4x^2)(lnx) + (1/x)(1/2x), you need to simplify the expression and differentiate further.

Let's simplify the expression first:

(1/4x^2)(lnx) + (1/x)(1/2x) can be written as (1/4x^2)(lnx) + 1/(2x^2)

Now, let's differentiate each term separately:

d/dx [(1/4x^2)(lnx)] = (1/4x^2)(d/dx(lnx)) + (d/dx(1/4x^2))(lnx)

To differentiate ln(x), we apply the chain rule: d/dx(lnx) = 1/x

So, (1/4x^2)(d/dx(lnx)) becomes (1/4x^2)(1/x) = 1/(4x^3)

Now, differentiating (d/dx(1/4x^2))(lnx):

d/dx(1/4x^2) = -2/(4x^3) = -1/(2x^3)

So, (d/dx(1/4x^2))(lnx) becomes (-1/(2x^3))(lnx) = -lnx/(2x^3)

Now, let's differentiate the second term: 1/(2x^2)

d/dx(1/(2x^2)) = -2/(2x^3) = -1/(x^3)

Now, we can put everything together:

dy/dx = (1/4x^2)(lnx) + 1/(2x^2) = 1/(4x^3) - lnx/(2x^3) - 1/(x^3)

Therefore, the derivative of y = sqrt(x)^(1/x) is dy/dx = 1/(4x^3) - lnx/(2x^3) - 1/(x^3).