full work out with explanation of this problem please we're working with derivatives and i don't understand it

given h(x)= cotx/x find h'(pi/2)

h = f/g so

h' = (f'g - fg')/g^2

h'(x) = [-x csc^2(x) + cot(x)]/x^2

h'(pi/2) = [-pi/2 * 1 + 0]/(pi^2 / 4)
= - 2/pi

To find the derivative of h(x) = cot(x)/x and evaluate it at x = π/2, we will need to use the quotient rule of differentiation. The quotient rule states that if we have a function in the form f(x) = g(x) / h(x), then its derivative is given by:

f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2

Let's break down the problem step by step:

Step 1: Write the given function.
h(x) = cot(x)/x

Step 2: Differentiate the numerator and denominator separately.
We need to differentiate cot(x) with respect to x. The derivative of cot(x) is -csc^2(x).
The derivative of x with respect to x is simply 1.

Step 3: Apply the quotient rule.
Using the quotient rule, we have:
h'(x) = (1 * cot(x) - x * (-csc^2(x))) / (x^2)
= (cot(x) + x * csc^2(x)) / (x^2)

Step 4: Evaluate at x = π/2.
To find h'(π/2), we substitute x = π/2 into h'(x):
h'(π/2) = (cot(π/2) + (π/2) * csc^2(π/2)) / ((π/2)^2)

At this step, we need to evaluate cot(π/2) and csc^2(π/2):
cot(π/2) = cos(π/2)/sin(π/2) = 0/1 = 0
csc(π/2) = 1/sin(π/2) = 1/1 = 1

Now, substituting these values back into h'(π/2):
h'(π/2) = (0 + (π/2) * 1) / ((π/2)^2)
= π/2 / (π^2/4)
= 2π / π^2
= 2/π

Therefore, h'(π/2) = 2/π.