Difference between revisions of "Derivative of arccos"

Latest revision as of 07:29, 8 June 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos}(z) = -\dfrac{1}{\sqrt{1-z^2}},$$ where $\mathrm{arccos}$ denotes the inverse cosine function.

Proof

If $\theta=\mathrm{arccos}(z)$ then $\cos(\theta)=z$. Now use implicit differentiation with respect to $z$ to get $$-\sin(\theta)\theta'=1.$$ The following image shows that $\sin(\mathrm{arccos}(z))=\sqrt{1-z^2}$:

Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccos}(z) = -\dfrac{1}{\sin(\mathrm{arccos}(z))} = -\dfrac{1}{\sqrt{1-z^2}}.█$$

References