higher derivative

Question

f(x) = x-1÷x+1 find the f(X)ⁿ

Answer 1

I'm assuming you mean (x-1)/(x+1).

 

f'(x) = [(x+1)*1 - (x-1)*1] / (x+1)^2 which simplifies down to 2/(x+1)^2 = 2(x+1)^-2

f"(x) = -4(x+1)^-3  *1 = =4(x+1)^-3

f'''(x) = 12(x+1)^-4

f''''(x) = -48(x+1)^-5

 

So let's see if we can generalize this. The (x+1) doesn't change.

 

 The exponent is one more than the derivative....but negative. So if you have the nth derivative, you get -(n+1) as the exponent. 

 

The coefficient alternates between positive and negative. We need it positive for when you have an odd numbered derivative and negative when you have an even derivative. So, (-1)^(n+1) will take care of that.

 

The coefficient also is formed by 2*2*3*4*5*..... which is related to factorial. f''''(x) has a coefficient of 48 (ignoring the negative since that's already taken care of. 48 = 2*2*3*4 which equals 2*4! . Let's see if that works for f'''(x). The coefficient is 12 so 2*3! works. So for the nth derivative, you get 2*n!

 

Putting it all togethter we get

the nth derivative of f(x) = (-1)^(n+1) * (2*n!)(x+1)^-(n+1).

 

Hope this makes sense to you.