5.2 The Natural logarithmic function

Differentiate the function
problem # 13: f(x)= sqrt(x) Inx

just use the product rule:

If y = fg
y' = f'g + fg'

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

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

(d/dx) [u(x) * v(x)] = u'(x) * v(x) + u(x) * v'(x)

In this case, we can let u(x) = sqrt(x) and v(x) = ln(x). We need to find the derivatives of u(x) and v(x), which we can do as follows:

1. Derivative of u(x) = sqrt(x):
To find the derivative of sqrt(x), we can use the power rule for differentiation. The power rule states that if we have a function f(x) = x^n, then the derivative is given by:

d/dx [x^n] = n * x^(n-1)

In this case, n = 1/2 (since sqrt(x) is the same as x^(1/2)). Therefore, the derivative of sqrt(x) is:

d/dx [sqrt(x)] = (1/2) * x^(-1/2)

2. Derivative of v(x) = ln(x):
To find the derivative of ln(x), we can use the chain rule. The chain rule states that if we have a function f(g(x)), where g(x) is an inner function and f(u) is an outer function, then the derivative is given by:

d/dx [f(g(x))] = f'(g(x)) * g'(x)

In this case, f(u) = ln(u) and g(x) = x. The derivative of ln(u) with respect to u is simply 1/u. Therefore, the derivative of ln(x) is:

d/dx [ln(x)] = 1/x

Now, let's use the product rule to find the derivative of f(x) = sqrt(x) ln(x):

f'(x) = [d/dx (sqrt(x))] * ln(x) + sqrt(x) * [d/dx (ln(x))]

Substituting in the derivatives we found earlier:

f'(x) = [(1/2) * x^(-1/2)] * ln(x) + sqrt(x) * (1/x)

Simplifying this expression:

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

So, the derivative of f(x) = sqrt(x) ln(x) is (1/2) * x^(-1/2) * ln(x) + sqrt(x)/x.