How do you differentiate # g(x) = sqrtarcsin(x^2-1) #?

1 Answer
Cesareo R.
May 24, 2016

#(dy)/(dx) =x/(sqrt(arcsin(x^2-1))sqrt[1 - (1 - x^2)^2])#

Explanation:

Define #y = sqrt(arcsin(x^2-1))# and make
#y^2=arcsin(x^2-1)->x^2-1=sin(y^2)#. Now following with
#f(x,y)=x^2-sin(y^2) = 1#
We know that #(dy)/(dx) = -f_x/(f_y) = (2x)/(2y cos(y^2))#. Now, substituting for #y = sqrt(arcsin(x^2-1))# we have
#(dy)/(dx) = x/(sqrt(arcsin(x^2-1))cos(arcsin(x^2-1))#
Simplifying
#(dy)/(dx) =x/(sqrt(arcsin(x^2-1))sqrt[1 - (1 - x^2)^2])#

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