Question

How do you find the derivative of `1-v^2/c^2`?

Apparently it's `(-2v)/c^2`, but how?
My steps:

(Quotient Rule)
#(c^2 * 2v - v^2 * 2c)/(c^2)^2 = -(2v)/c^2 * (c^2 +cv)/c^2#

What did I do wrong do get an answer off by a factor of `(c^2 +cv)/c^2`?

Answer 1

Either `c` is a constant or you are finding a partial derivative with respect to `v` (in which case `c` is treated like a constant).

Explanation:

You don't mention what the variable is here are some possibilities:

Find `d/(dv)(1-v^2/c^2)` where `c` is a constant.

In this case, the function of interest is `f(v) = 1-v^2/c^2` where `c` is a constant., and derivative of `c` is `0`, not `2c`

If you really want to use the quotient rule, it goes like this:

`d/(dv)(1-v^2/c^2) = 0- (2v(c^2) - v^2 (0))/(c^2)^2`

`= -(2vc^2)/c^4 = -(2v)/c^2`

OR

The function of interest is

`f(v,c) = 1-v^2/c^2`

and you are looking for `(delf)/(delv)`.

(Which is also denoted `f_v(v,c)` or `D_v(f(v,c))` or `D_1(f)` and probably some others I've left out.)

For the partial derivative with respect to `v`, we treat `c` as a constant, so we get the same result as above.

Answer 2

It looks like the variable is `v`, in fact:

`d/(dv) (1-v^2/c^2) = -(2v)/c^2`

if `c` is constant