Derivative of Arccosine Function

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Theorem

Let $x \in \R$ be a real number such that $-1 < x < 1$.

Let $\arccos x$ be the arccosine of $x$.


Then:

$\map {D_x} {\arccos x} = \dfrac {-1} {\sqrt {1 - x^2}}$


Corollary

$\dfrac {\mathrm d \left({\arccos \left({\frac x a}\right)}\right)} {\mathrm d x} = \dfrac {-1} {\sqrt {a^2 - x^2}}$


Proof

Let $y = \arccos x$ where $-1 < x < 1$.

Then:

$x = \cos y$

Then from Derivative of Cosine Function:

$\dfrac {\d x} {\d y} = -\sin y$

Hence from Derivative of Inverse Function:

$\dfrac {\d y} {\d x} = \dfrac {-1} {\sin y}$

From Sum of Squares of Sine and Cosine, we have:

$\cos^2 y + \sin^2 y = 1 \implies \sin y = \pm \sqrt {1 - \cos^2 y}$

Now $\sin y \ge 0$ on the range of $\arccos x$, that is, for $y \in \closedint 0 \pi$.

Thus it follows that we need to take the positive root of $\sqrt {1 - \cos^2 y}$.

So:

$\dfrac {\d y} {\d x} = \frac {-1} {\sqrt {1 - \cos^2 y} }$

and hence the result.

$\blacksquare$


Also see


Sources