How do you differentiate #(y^2+1)^2-x=0#?

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Answer 1 · 2018-04-10T06:27:09

The implicit derivative #dy/dx# is #1/(4y(y^2+1))#.

First, rearrange a little:

#(y^2+1)^2-x=0#

#(y^2+1)^2=x#

Now, take the implicit derivative (which basically means treat #y# as a function of #x#):

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

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

#2(y^2+1)^(2-1)*d/dx[y^2-1]=1#

#2(y^2+1)^(1)*d/dx[y^2-1]=1#

#2(y^2+1)*d/dx[y^2-1]=1#

#d/dx[y^2-1]=1/(2(y^2+1))#

#dy/dx*2y=1/(2(y^2+1))#

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

#dy/dx=1/(4y(y^2+1))#

That's the derivative. Hope this helped!