# How do you find the derivative of the function: y=arccos(arcsin x)?

Mar 2, 2016

d/dxarccos(arcsin(x))=-1/(sqrt(1-x^2)sqrt(1-arcsin^2(x))

#### Explanation:

Noting that
$\frac{d}{\mathrm{dx}} \arccos \left(x\right) = - \frac{1}{\sqrt{1 - {x}^{2}}}$
and
$\frac{d}{\mathrm{dx}} \arcsin \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$
(derivations are included at the bottom)

Applying the chain rule gives us

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

$= - \frac{\frac{1}{\sqrt{1 - {x}^{2}}}}{\sqrt{1 - {\arcsin}^{2} \left(x\right)}}$

=-1/(sqrt(1-x^2)sqrt(1-arcsin^2(x))

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To derive the derivatives used above, we can use implicit differentiation.

Let $y = \arccos \left(x\right)$

$\implies \cos \left(y\right) = x$

$\implies - \sin \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sin} \left(y\right)$

$= - \frac{1}{\sin} \left(\arccos \left(x\right)\right)$

$= - \frac{1}{\sqrt{1 - {x}^{2}}}$

(Try drawing a right triangle where $\cos \left(\theta\right) = x$ and calculate $\sin \left(\theta\right)$ for the final step)

Let $y = \arcsin \left(x\right)$

$\implies \sin \left(y\right) = x$

$\implies \cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(y\right)$

$= \frac{1}{\cos} \left(\arcsin \left(x\right)\right)$

$= \frac{1}{\sqrt{1 - {x}^{2}}}$