Hi
For all x in (-r,r)
-f'(-x) = f'(x)
(f(-x))' = (f(x))'
Integrating gives
f(-x) = f(x) + K
To find K we take x=0
f(0) = f(0) + K
K = 0
And finally
f(-x) = f(x)
using differentiation, i want to show
f is even on (-r,r) if and only if f' is odd on (-r,r) and also odd-even relation...
in proving f=>f', differentiating the eqation seems sufficient, but
i dont know how to do the converse, proving f'=>f using differentiation...
Use the chain rule: for any function f(x), with u= -x. That is, .
If f is an even function, then f(-x)= f(x) so and that previous equation becomes f'(x)= -f'(-x) which is the same as -f'(x)= f'(-x) showing that f'(x) is an odd function.
If f is an odd function, then f(-x)= -f(x) so and that previous equation becomes -f'(x)= -f'(-x) which is the same as f'(x)= f'(-x) showing that f'(x) is an even function.