Thread: nth derivative

Find the nth derivative of 1/(ax + b)^1/2

I got ((2n-1)!/2^n+1) * (-a)^n * (ax+b)^(-(1+2n)/2)

but the answer in the back of the book was

(2 * (2^2n)!) / (((-1)^n) * (a^n) * ((2n)!) * (ax+b)^-(n + 0.5))

Not sure whether ive got it wrong or it's just in a different form... never done nth derivatives before though, so thought I'd check

Originally Posted by kinetix2006

Find the nth derivative of 1/(ax + b)^1/2

I got ((2n-1)!/2^n+1) * (-a)^n * (ax+b)^(-(1+2n)/2)

Can I assume that your "2^n+1" should be "2^(n+1)", not "(2^n)+ 1"?

but the answer in the back of the book was

(2 * (2^2n)!) / (((-1)^n) * (a^n) * ((2n)!) * (ax+b)^-(n + 0.5))

Not sure whether ive got it wrong or it's just in a different form... never done nth derivatives before though, so thought I'd check

It would help if you showed HOW you got that. Your answer is clearly incorrect. For example, for n= 1 it gives (2-1)!/2^2 (-a)^1(ax+b)^(-(1+2)/2)= (1/4)(-a)(ax+b)^(-3/2) while the correct first derivative is (1/2)(-a)(ax+b)^(-3/2). For n= 3, your formula gives -(15/2)a^3(ax+ b)^(-7/2)
while the correct third derivative is -(15/8)a^3(ax+b)^{-7/2}.

Interestingly, your formula gives the correct second derivative!

(-a)^n= (-1)^n a^n so your formula is only wrong in that (2n-1)!/2^(2n+1) part. (If I assume you really mean (2^n)+ 1, it's worse!)

Yeah I did mean 2^(n+1)

I got to (z/2^n) * (-a)^n * (ax+b)^(-(1 + 2n)/2)

where z is basically (2n - 1)! but only the odd numbers but im not sure how i get that algebraically

I only tried (2n-1)!/2^(2n+1) for the 2nd derivative which is why it worked for that lol.

I can't see how it relates to the answer in the back of the book either as it doesn't seem to work