What is the derivative of f(x)=(x^2)/(1-x^2)? please help?

Question

Please and thankyou

Answer 1

This derivative requires the quotient rule.
dy/dx=((1-x^2)(2x)-(x^2)(-2x))/(1-x^2)^2
dy/dx=((2x-2x^3)-(-2x^3))/(1-x^2)^2
dy/dx=2x/(1-x^2)^2

Answer 2

The derivative is found based on a limit, either as some number (s) to x, or some other number h as it moves to zero. Using the limit h --> 0 is a little easier (in the end, substitute zero for h)

f'(x) = [lim h --> 0] ( f(x + h) - f(x) ) / h
You plug your formula into 'x' and solve

f'(x) = [lim h --> 0] ( [((x + h)^2)/(1-(x + h)^2)] - [(x^2)/(1-x^2)] ) / h

Solve this all down, then put zero (0) for h at the end.

Answer 3

3

Answer 4

You have to use the quotient rule.
[ (1-x^2)*2x-x^2*(-2x) ]/(1-x^2)^2

Answer 5

you just need to do the quotient rule on this one... the easiet way to remember it is this little jingle....... low d high minus high d low square the bottom and away we go...which means take the (bottom number times the der. of the high number) minus the (high number times the der. of the bottom num.) then put it all over the square of the bottom number ...then simplify and you got it.

Answer 6

d f(x)/ d(x) = 2x (1-x^2) + 2x^3 / {(1-x^2)} ^2

Answer 7

(2x)/(1-x^2) + 4x^3 * [1-x^2]^-2

Answer 8

{ [(1-x^2) * 2x] - [x^2 * (-2x)]} / { [1-x^2]^2}

so when its (a/b) you get:

[b * da/dx] - [a *db/dx] / b^2

Answer 9

say what

Answer 10

What grade are you!!!??