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)#