find the nth derivative of 2/(x+1)
n=1; f'=-2/(x+1)^2
n=2; f"=2*2/(x+1)^3
nth = (-2)^n * 1/(x+1)^(n+1)
check that.
that is only true for the first three.
To find the nth derivative of a function, we can use the power rule for derivatives and also apply the chain rule if necessary. Let's start by taking the derivatives step by step:
Given function: f(x) = 2/(x+1)
First derivative (n = 1):
f'(x) = 2 * d/dx[(x+1)^(-1)]
Applying the power rule for derivatives (d/dx[x^n] = n * x^(n-1)) and the chain rule (d/dx[f(g(x))] = f'(g(x)) * g'(x)):
f'(x) = 2 * (-1) * (x+1)^(-2) * d/dx[x+1]
= -2/(x+1)^2
Second derivative (n = 2):
f''(x) = d/dx[-2/(x+1)^2]
Using the power rule and chain rule again:
f''(x) = -2 * d/dx[(x+1)^(-2)]
= -2 * (-2) * (x+1)^(-3) * d/dx[x+1]
= 4/(x+1)^3
Therefore, the second derivative of the function f(x) = 2/(x+1) is f''(x) = 4/(x+1)^3.
To find higher derivatives, we can continue applying the power rule and chain rule in a similar manner. Each derivative will introduce an additional factor of (-1) and raise the power of (x+1) in the denominator by one.
In general, the nth derivative (where n > 1) of f(x) = 2/(x+1) can be written as:
f^(n)(x) = (-1)^(n) * (n!) * (2/(x+1)^(n+1))
where "!" denotes the factorial function representing the product of all positive integers up to n.