nth derivative of a function?

Question

How do you find the nth derivative of a standard function in algebra?an example would be great

Answer 1

Assuming you want a general formula for the nth derivative, usually the only way of obtaining it is to take multiple derivatives until you notice the pattern that's going on. Of course, for a polynomial, after you reach some particular value of n, every derivative is 0 as mentioned above.



Some simple examples (the 0th derivative starts each list, then increases by one a few times, then finally notes the nth derivative):

x^b

b*x^(b-1)

b*(b-1)*x^(b-2)

b*(b-1)*(b-2)*x^(b-3)

...

so for n<=b, nth derivative is b!/(b-n)! * x^(b-n)

and for n>b, nth derivative is 0



e^(bx)

be^(bx)

b^2e^(bx)

b^3e^(bx)

....

b^n*e^(bx)



For finding Taylor series, you will need to find a formula for the nth derivative evaluated at a certain x-value.

Answer 2

f(x) = ax^b



1st derivative = abx^(b-1)

2nd derivative = ab(b-1)x^(b-2)

3rd derivative = ab(b-1)(b-2)x^(b-3)

.

.

.

nth derivative =

ab(b-1)(b-2) ... (b-(n-1))x^(b-n)



At some point the power of x will become zero, and the next derivative will be zero.



Not sure if this is what you were looking for, but hope it helps.

Answer 3

each and each successive by-product (as a fragment in case you used quotient rule) could have a element of e^x contained in the numerator and denominator, so once you sq. the bottom, between the e^x's will continuously cancel out. The numerator is going from a million-x to -2+x to three-x, etc. This trend should be generalized via (-a million)^n cases (x - n) --------------------------- e^x For n=a million, the numerator is -(x - a million), or a million-x. For n=2, the num is +(x - 2), or x-2. For n=3, num is -(x - 3), or 3-x. etc.

Answer 4

just keep taking derivatives of your derivatives until you reach the nth derviative. if you have a standard algebraic function (read polynomial), eventually you will get zero as your answer.



if the degree of your polymonial is k, then the k-th derivative will be a constant (it is actually k!) and the (k+1)-st derivative will be zero