How do you find #dy/dx# given #x=y^2+1#?

Question

1024 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-28T12:44:23

#dy/dx = +-1/(2sqrt(x-1))#

Differentiating both sides of the equation with respect to #x# we have:

#1 = 2ydy/dx#

#dy/dx = 1/(2y)#

Now, as #x = y^2+1 > 1# we have #y= +-sqrt(x-1)#

and:

#dy/dx = +-1/(2sqrt(x-1))#

Answer 2 · 2017-02-28T12:44:40

#dy/dx = +-1/(2sqrt(x-1)#

#x=y^2+1#

#y^2 = x-1#

#dy/dx = 1/(2y)#

Since #y^2 = x-1# then #y = +- sqrt(x-1)#

#:. dy/dx = +-1/(2sqrt(x-1)#