Stuck on a simple calculus problem. Can you help?

f(x) = x^2 + 1/1 -x^2

f '(x) = ((x^2 + 1)' ( 1-x^2) -(1 - x^2)' (x^2 + 1))/(1 - x^2)^2

= (2x(1 -x^2) -(-2x) (x^2 + 1))/(1 -x^2)^2

= (2x -2x^3 + 2x(x^2 + 1))/(1 - x^2)^2

= (2x -2x^3 + 2x^3 + 2x)/(1 -x^2)^2

= 4x/(1 -x^2)^2

Conclusion : Your answer is right.

Unless I am mistaken, I believe that (1-x^2)^2 is equal to (x^2-1)^2, so you got the right answer!

(1-x^2)^2 = 1 - 2x^2 + x^4 = x^4 - 2x^2 +1

(x^2-1)^2 = x^4 -2x^2 + 1

Go ahead and foil out (x^2 - 1) ^ 2 and (1 - x^2)^2 and see how the answers compare - your answer is equivalent to the one they came up with.

Hi Mitch.

I looked into both answers and apparently if you get d2/dx they both work out to be the same.

Therefore, I wouldn't worry about it.

I didn't take a look at your calculations. Hopefully you started out with the quotient rule.

another thing why it doesn't matter is because you are anyway squaring your denominator which cancels out the whole notion of getting a completely different answer.

Hope this helped.

Yea, you are right, just plug in a number for your answer and the "correct" answer and you'll get the same thing

Your answer is correct.

4x / (1-x^2)^2 = 4x / (x^2-1)^2

*sticks tongue out

now if you look at your answer, it is pretty much the same just switched around.

ur answer is the correct answer they just worded theres another way

LOL "SIMPLE" calculus problem. ROFLMAO

Seriously, it looks like a factoring issue to me.

d/dx (u/v)= (vdu/dx-udv/dx)/v^2

so f '(x)= ((1-x^2)(2x)-(x^2+1)(-2x))/(1-x^2)^2...

just simplify this.