How do you find the derivative of the inverse of #y = x^2 + 1#?

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Answer 1 · 2016-05-02T10:40:51

#y'=1/(2y)=1/(+-2sqrt(x-1)#

The given function is

#y=x^2+1#

The inverse is

#x=y^2+1#

Differentiate both sides of the equation with respect to x

#d/dx(x)=d/dx(y^2+1)#

#1=2yy'+0#

#y'=1/(2y)#

but #y=+-sqrt(x-1)#

therefore

#y'=1/(2*(+-sqrt(x-1)))=1/(+-2*sqrt(x-1)#

God bless....I hope the explanation is useful.