Finding Derivative of (1-x^2)^-1/4: Solving for 0

[sobek]

How would I find the derivative of (1-x^2)^-1/4?
y = 1 + x - x^2 - x^4

I found the derivative to be 1 - 2x - 4x^3, but I need help solving it for 0. Thanks for your time.

 

[HallsofIvy]

How would I find the derivative of (1-x^2)^-1/4?
y = 1 + x - x^2 - x^4

I found the derivative to be 1 - 2x - 4x^3, but I need help solving it for 0. Thanks for your time.

? Are you talking about two different functions to differentiate? Yes, the derivative of y= 1+ x- x^2- x^4 is 1- 2x- 4x^3. There is a "cubic formula" but it is very complicated. If 1- 2x- 4x^3 does not have rational roots, there is not going to be any simple solution. If there are rational roots then they must be $\pm 1$, $\pm 1/2$, or $\pm 1/4$. Do any of those satisfy this equation? If so you can divide by x- that number to reduce the rest to a quadratic.

To differentiate (1- x^2)^(-1/4), use the chain rule. The derivative of ( )^(-1/4) is (-1/4)( )^(-1/4-1)= (-1/4( )^(-5/4) times the derivative of what ever is in the ( ). Here, that is 1- x^2 which has derivative -2x. The derivative of (1- x^2)^(-1/4) is (-1/4)(1- x^2)^(-5/4)(-2x)= (1/2)x(1- x^2)^(-5/4).