How do you find the derivative of the function: y=arccos(arcsinx)?

1 Answer
sente
Mar 2, 2016

ddxarccos(arcsin(x))=−1√1−x2√1−arcsin2(x)

Explanation:

Noting that
ddxarccos(x)=−1√1−x2
and
ddxarcsin(x)=1√1−x2
(derivations are included at the bottom)

Applying the chain rule gives us

ddxarccos(arcsin(x))=−1√1−arcsin2(x)(ddxarcsin(x))

=−1√1−x2√1−arcsin2(x)

=−1√1−x2√1−arcsin2(x)

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

Let y=arccos(x)

⇒cos(y)=x

⇒−sin(y)dydx=1

⇒dydx=−1sin(y)

=−1sin(arccos(x))

=−1√1−x2

(Try drawing a right triangle where cos(θ)=x and calculate sin(θ) for the final step)

Let y=arcsin(x)

⇒sin(y)=x

⇒cos(y)dydx=1

⇒dydx=1cos(y)

=1cos(arcsin(x))

=1√1−x2

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