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?

2 Answers
Jul 26, 2017

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.

Jul 26, 2017

It looks like the variable is v, in fact:

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

if c is constant