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

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Answer 1 · 2016-03-02T09:08:07

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

Noting that
$d/dx arccos(x) = -1/sqrt(1-x^2)$
and
$d/dx arcsin(x) = 1/sqrt(1-x^2)$
(derivations are included at the bottom)

Applying the chain rule gives us

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

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

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

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

Let $y = arccos(x)$

$=> cos(y) = x$

$=> -sin(y)dy/dx = 1$

$=> dy/dx = -1/sin(y)$

$=-1/sin(arccos(x))$

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

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

Let $y = arcsin(x)$

$=> sin(y) = x$

$=> cos(y)dy/dx = 1$

$=> dy/dx = 1/cos(y)$

$=1/cos(arcsin(x))$

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

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