Because
is its own derivative, our anti-derivative for the function
will probably be very similar to the original function. To check our first anti-derivate candidate
we have to use the chain rule,
, where
and
. Then
and
, so we calculate:
![{\displaystyle \displaystyle (e^{3-x})'=(G(H(x)))'=G'(H(x))H'(x)=(e^{3-x})(-1)=-e^{3-x}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/66d5605a5fa9e5ac8e5429e9003cf7e297b4cbd5)
Hence our candidate almost works, we just have to bring the minus sign over:
. Using the chain rule to double check, we find that, indeed,
![{\displaystyle \displaystyle F'(x)=(-e^{3-x})'=(-e^{3-x})(-1)=e^{3-x}=f(x).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e8f66d619e02be3cba6e19d51009a05f5f5861e2)
This is one anti-derivative. To find another one, we can simply add a constant to the anti-derivative shown above, say
. When we differentiate, the constant will disappear, giving us the same derivative as before.