7-6-29 derivative of inverse secant
X
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CC
When doing this problem, we have the variable in the base and the
exponent.
So we're going to take E to the lane of X to the -5 / l and X.
The limit as X goes to zero from the right hand side.
So we can bring the E out as a base and just look at the limit
as X goes to zero from the right of the exponent.
We're going to use the power rule to bring down that -5 lane
X times lane X.
The lane X's here are actually going to cancel, so we're going
to get E limit as X goes to zero from the right of -5 There is no
X there.
So we can now see that that's really just -5.
So we get E to the -5 or 1 / E to the 5th.