Computing Derivatives Exercises

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

Answer

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.