Verified answer

To answer this, you need to use the product rule.

I will use "sqrt" to mean "square root."

The product rule is "derivative of the first times the second plus the derivative of the second times the first." In symbols,

(f(x)g(x))' = f '(x)g(x) + f(x)g'(x)

x*sqrt(ln x) = x*(ln x)^(1/2) (just rewriting here)

Using the product rule:

1*(ln x)^(1/2) + (1/2)*(ln x)^(-1/2)*(1/x)*x

Simplifying...

f ' (x) = (1 + 2(ln x)) / 2*sqrt(ln x)

Or another form...

f '(x) = sqrt(ln x) + 1 / (2*sqrt(ln x))

Re-writing the original equation as: f(x) = [x^0.5]*[lnx]^2 Using the product rule, which states: d/dx(uv) = u*dv/dx + v*du/dx Let: u = x^0.5 du/dx = 0.5*x^(-0.5) v = [lnx]^2 You can make use of the chain rule to determine dv/dx. The chain rule states: dv/dx = dv/dw*dw/dx Let: w = lnx dw/dx = 1/x v = [lnx]^2 = w^2 dv/dw = 2w Therefore: dv/dx = dv/dw*dw/dx dv/dx = 2w*(1/x) dv/dx = 2w/x Substituting w back in, we get: dv/dx = 2(lnx)/x So the first derivative is f'(x) = u*dv/dx + v*du/dx f'(x) = (x^0.5)*2[lnx]/x + ([lnx]^2)*0.5*x^(-0.5) f'(x) = (x^-0.5)*2[lnx] + 0.5x^(-0.5)*([lnx]^2) f'(x) = (x^-0.5)*lnx*(2 + 0.5lnx) f'(x) = lnx*(2 + 0.5lnx)/(x^0.5) Multiplying by 2/2, which is the same as multiplying by 1, we get: f'(x) = (2/2)lnx*(2 + 0.5lnx)/(x^0.5) f'(x) = lnx*(4 + lnx)/(2x^0.5) This matches your answer. To find the second derivative (f''(x)), we use the quotient rule. The quotient rule states, if given: a(x) = g(x)/h(x) and you want to find a'(x), then a'(x) = [h(x)g'(x) - h'(x)g(x)]/[h(x)]^2 Let: a(x) = f'(x) a'(x) = f''(x) g(x) = lnx*(4 + lnx) g(x) = 4lnx + (lnx)^2 Now we take the derivative of g(x) to get g'(x). When we took the derivative of the original function we found that the derivative of (lnx)^2 was (2lnx)/x. Therefore: g'(x) = 4/x + (2lnx)/x g'(x) = 2(2 + lnx)/x h(x) = 2x^0.5 h'(x) = 1/x^0.5 Putting this together, we get: f''(x) = [h(x)g'(x) - h'(x)g(x)]/[h(x)]^2 f''(x) = [2x^0.5*2(2 + lnx)/x - (1/x^0.5)*lnx*(4 + lnx)]/[2x^0.5]^2 f''(x) = [2(4 + 2lnx)/x^0.5 - lnx*(4 + lnx)/x^0.5]/4x f''(x) = [8 + 4lnx - (4lnx + (lnx^2))]/4x^1.5 f''(x) = [8 + 4lnx - 4lnx - (lnx^2)]/4x^1.5 f''(x) = [8 - (lnx^2)]/4x^1.5

first derivative of f(x) is sqrt(lnx) + x*(1/x)*(1/(2sqrt(lnx)))

0r simplified as sqrt(lnx) + 1/(2sqrt(lnx))