? 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 [itex]\pm 1[/itex], [itex]\pm 1/2[/itex], or [itex]\pm 1/4[/itex]. Do any of those satisfy this equation? If so you can divide by x- that number to reduce the rest to a quadratic.sobek said: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.
The derivative of (1-x^2)^-1/4 is 0.
To solve for 0 in the derivative of (1-x^2)^-1/4, we set the derivative equal to 0 and solve for x. In this case, we get 1-x^2 = 0, which gives us x = ±1.
The derivative of (1-x^2)^-1/4 is equal to 0 because it is a constant function. The derivative of a constant function is always 0.
Solving for 0 in the derivative of (1-x^2)^-1/4 means finding the values of x that make the derivative equal to 0. These values are also known as the critical points of the function.
No, the power rule cannot be used directly to find the derivative of (1-x^2)^-1/4. Instead, the chain rule must be used in combination with the power rule to find the derivative of this function.