What is the derivative of # arcsin(x^4)#?

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Answer 1 · 2017-06-02T02:43:26

#dy/(dx)=(4x^3)/sqrt(1-x^8)#

Let #u=x^4# then #(du)/(dx)=4x^3#

Let #y=arcsin(u)#

Take the arcsin of both sides

#sin(y)=u#

Differentiate both sides with respect to #u#

#dy/(du)cos(y)=1#

Divide both sides by #cos(y)#

#dy/(du)=1/cos(y)#

recall that #Sin^2(y)+cos^2(y)=1#

So #cos(y)=sqrt(1-sin^2(y))#
NOTE: Take only the positive root

From above we said that #sin(y)=u# so we can write

#cos(y)=sqrt(1-u^2)#

So #dy/(du)=1/sqrt(1-u^2)#

Recall that #u=x^4#

So #dy/(du)=1/sqrt(1-(x^4)^2)#

Now using the chain rule

#dy/dx=dy/(du)(du)/(dx)#

#dy/dx=1/sqrt(1-x^8)(4x^3)#

#dy/dx=(4x^3)/(sqrt(1-x^8))#