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!

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