Calculus Problem: Derive (1) Explicitly

[cosmology]

how do you derive (1)?please work it out explicitly

 

[HallsofIvy]

What you have is
[texJ(q^2)= -/frac{1}{2}ln^2(\frac{-q^2}{\lambda^2})+ 2ln(\frac{-q^2}{\lambda^2})+ \frac{\pi^2}{6}-\frac{5}{2}[/tex]
If q and $\lambda$ are real, then those logarithms are not defined.

 

[tacman]

To derive (1) explicitly, we can use the definition of a derivative. The derivative of a function f(x) at a point x=a is defined as the limit of the slope of the secant line passing through the points (a, f(a)) and (x, f(x)) as x approaches a. In other words, it is the instantaneous rate of change of the function at that point.

In this case, we want to find the derivative of the function f(x) = 1 at any point x=a. Using the definition, we can write the derivative as:

f'(a) = lim (x→a) (f(x) - f(a)) / (x - a)

Since f(x) = 1 for all values of x, we can substitute it into the equation above:

f'(a) = lim (x→a) (1 - 1) / (x - a)

Simplifying, we get:

f'(a) = lim (x→a) 0 / (x - a)

Since the limit of any constant over a variable is always 0, we have:

f'(a) = 0

Therefore, the derivative of the function f(x) = 1 at any point x=a is 0. This makes sense because the function f(x) = 1 is a horizontal line, meaning it has a constant slope of 0 at all points.

In summary, to derive (1) explicitly, we used the definition of a derivative and substituted in the given function to find that the derivative is equal to 0 at any point x=a.