Find the nth derivative of y=2x/(1-x^2)
(Hint: Consider -[(1/(x-1))+(1/(x+1))].)
so, did you even think about the hint?
2x/(1-x^2) = 2x/((1-x)(1+x)
= -[1/(x+1) + 1/(x-1)]
The nth derivative of 1/(x+1) and 1/(x-1) is easy to figure, right?
Or how do I find the formula for this?
huh? Let's take f=1/u. (Let u=x+1 or u=x-1)
f = 1/u = u^-1
f' = -1 u-2
f" = (-1)(-2)u-3
...
fn = (-1)nn! u-(n+1)
To find the nth derivative of y = 2x/(1 - x^2), we can use the Quotient Rule repeatedly.
The Quotient Rule states that if we have a function u(x) divided by v(x), where both u(x) and v(x) are differentiable functions, then the derivative of this quotient is given by:
d/dx(u(x)/v(x)) = (v(x) * u'(x) - u(x) * v'(x)) / [v(x)]^2
Let's start by rewriting the function y = 2x/(1 - x^2) using algebraic manipulation and simplify it:
y = 2x / (1 - x^2)
= 2x / [(1 - x)(1 + x)]
= 2x / [1 - x^2]
= 2x * [1 - x^2]^-1
Now let's use the Quotient Rule to find the first derivative:
d/dx (2x * [1 - x^2]^-1) = [(1 - x^2)^-1 * (2) - 2x * (-2x) * (1 - x^2)^-2] / [1 - x^2]^2
= [2/(1 - x^2) + 4x^2/(1 - x^2)^2] / [1 - x^2]^2
To find the second derivative, we need to take the derivative of this expression. Let's do it:
d/dx ([2/(1 - x^2) + 4x^2/(1 - x^2)^2] / [1 - x^2]^2)
To simplify this expression, we need to use the Quotient Rule again. Let's define u(x) = 2/(1 - x^2) + 4x^2/(1 - x^2)^2 and v(x) = [1 - x^2]^2. We can now apply the Quotient Rule:
(d/dx) [u(x) / v(x)] = (v(x) * u'(x) - u(x) * v'(x)) / [v(x)]^2
where u'(x) and v'(x) are the derivatives of u(x) and v(x) with respect to x.
Applying the Quotient Rule, we get:
d/dx ([2/(1 - x^2) + 4x^2/(1 - x^2)^2] / [1 - x^2]^2) = ([1 - x^2]^2 * [d/dx (2/(1 - x^2)) + d/dx (4x^2/(1 - x^2)^2)] - (2/(1 - x^2) + 4x^2/(1 - x^2)^2) * [d/dx ([1 - x^2]^2)]) / [[1 - x^2]^2]^2
Now we can simplify each term and continue this process for the nth derivative. However, since the initial hint suggests an alternative approach, let's explore that instead.
Consider the expression -[(1/(x - 1)) + (1/(x + 1))]. Simplifying this expression gives:
-[(1/(x - 1)) + (1/(x + 1))]
= -[(x + 1 + x - 1) / (x - 1)(x + 1)]
= -[2x / (x^2 - 1)]
= -[2x / (1 - x^2)]
Notice that this expression matches our original function y = 2x/(1 - x^2) with opposite sign. This implies that the nth derivative of y = 2x/(1 - x^2) is equal to the nth derivative of -[(1/(x - 1)) + (1/(x + 1))].
The advantage of using the expression -[(1/(x - 1)) + (1/(x + 1))] is that it is easier to differentiate using the Power Rule and basic algebraic manipulations.
Therefore, the nth derivative of y = 2x/(1 - x^2) can be found by differentiating -[(1/(x - 1)) + (1/(x + 1))] n times using the Power Rule.