Find the derivative of the function.
Answer
We already know cos(arccos x) = x,
so taking derivatives of both sides gives us,
-sin (arccos x)(arccos x)' = 1
which means
If we let θ = arccos x, then somewhere there's a right triangle with an angle s such that
That triangle looks something like this:
Using the Pythagorean theorem, the missing side of the triangle is
therefore
Then
Starting with tan(arctan x) = x,
we can take the derivative of each side to get,
sec2(arctan x)(arctan x)' = 1 so,
Let θ = arctan x. Since
somewhere there's a right triangle that looks like this:
The Pythagorean Theorem tells us that the missing side is
secant is the reciprocal of cosine, in other words, "hypotenuse over adjacent," therefore
Therefore
In the same way, we can find that
However, these derivatives don't seem to be used nearly as often as the other ones.
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