Find the derivative of 1/(1+(x-3)^2)

Steps would be appreciated.

if you let u = 1 + (x-3)^2 then

y = 1/u

The chain rule says that

y' = -1/u^2 u'
= -1/(1+(x-3)^2)^2 * (2(x-3))

To find the derivative of the given function, we can use the quotient rule.

The quotient rule states that for functions u(x) and v(x), the derivative of their quotient, f(x) = u(x)/v(x), is given by:

f'(x) = (u'(x)v(x) - v'(x)u(x))/[v(x)]^2

Let's apply the quotient rule to find the derivative of 1/(1+(x-3)^2).

Step 1: Identify the numerator and the denominator of the function.
Numerator: 1
Denominator: 1+(x-3)^2

Step 2: Find the derivative of the numerator.
The derivative of a constant (1) is always 0.

Step 3: Find the derivative of the denominator.
The numerator of the denominator is 1, and the derivative of a constant is always 0.
The denominator can be written as (1+(x-3)^2) = (x-3)^2 + 1, which is a sum of two functions.
To find its derivative, we can use the chain rule.

Let f(x) = (x-3)^2 + 1.
Applying the chain rule, the derivative of f(x) is 2(x-3) * (d/dx)(x-3) = 2(x-3).

Step 4: Apply the quotient rule.
Now, using the quotient rule, we have:

f'(x) = (0*(1+(x-3)^2) - (2(x-3))*(1))/[(1+(x-3)^2)]^2
= (-2(x-3))/[(1+(x-3)^2)]^2

Thus, the derivative of 1/(1+(x-3)^2) is (-2(x-3))/[(1+(x-3)^2)]^2.