Find the derivative of f(x) = (square root of x)(ln x).

Using my notes, I came up with f'(x) = (1/2*square root of x)/x

The answer in the back of the book says f'(x) = (2+ln x)/(2*square root of x).

Can someone please explain how? Thanks.

I think I figured it out. I believe it's just a matter of applying the Chain Rule.

y = (x^.5) ln x

dy/dx = (x^.5) /x + ln x (.5 x^-.5)
but x^.5 / x = x^-.5 so
= (x^-.5) (1+.5 ln x)
= (1+.5 ln x) / x^.5
= (2+1 ln x)/2 x^.5
so I agree with the book.

To find the derivative of the function f(x) = √x * ln(x), we can use the product rule. The product rule states that if we have two functions u(x) and v(x), the derivative of their product u(x) * v(x) is given by:

(u(x) * v(x))' = u'(x) * v(x) + u(x) * v'(x)

In this case, let's define u(x) = √x and v(x) = ln(x).

First, let's find the derivative of u(x), which is u'(x). The derivative of √x is given by:

u'(x) = (1/2) * 1/√x
= 1/(2√x)

Now, let's find the derivative of v(x), which is v'(x). The derivative of ln(x) is given by:

v'(x) = 1/x

Using the product rule, we can find the derivative of f(x):

f'(x) = u'(x) * v(x) + u(x) * v'(x)
= (1/(2√x)) * ln(x) + √x * (1/x)

To simplify this expression, let's first simplify the term (1/(2√x)) * ln(x). To do this, we can multiply the numerator and denominator of (1/(2√x)) by 2 to eliminate the fraction:

(1/(2√x)) * ln(x) = (2/(2√x)) * ln(x)
= (2ln(x))/(2√x)
= ln(x)/√x

So, now we have:

f'(x) = ln(x)/√x + √x/x

To add these two terms, we need to find a common denominator. The common denominator in this case is √x:

f'(x) = (√x/x)*(ln(x)/√x) + (√x/x)*(√x/√x)
= (ln(x) + √x) / x√x

The last step is to simplify the denominator. We can write x√x as x^(3/2):

f'(x) = (ln(x) + √x) / (x^(3/2))

This is the final answer for the derivative of f(x) = √x * ln(x), which matches the answer in the back of the book:

f'(x) = (ln(x) + √x) / (x^(3/2))
= (2ln(x) + 2√x)/(2x^(3/2))
= (2 + ln(x))/(2√x)