The Nth Derivative of a Function (General Form)

[RocketSurgery]

1. Question: Find the nth derivative{y^(n)} for the function y=(x+5)^(1/2)

I'm teaching myself calculus and so far I have been doing well getting through two workbooks I bought. I was going through a section in Higher Order Derivatives and Implicit Differentiation in the book "Schaum's 3000 Solved Problems in Calculus". I am well aware of how to get the 1st 2nd and 3rd derivative and so on... but I'm unable to figure out how to find the general form of this equation.

2. Attempt at a solution: For a simpler equation of the same time where it asked me to find a general form for the nth derivative I used the equation, http://upload.wikimedia.org/math/2/c/b/2cb203daf11bd73b58be0fd89bee1d48.pngBtw is this type of question normal for a 1st yr calculus course? I've never seen one like it on calc 1 webpages and workbooks.

 

[StatusX]

That formula isn't quite applicable to this problem, since you can only take the factorial of integers. There is a generalization of the factorial to non-integers which makes that true for all k, but it's a little complicated, and not worth the effort for this problem (unless you're interested).

Do you know the derivation of that formula? You can use induction to prove it, but it's simple to see where it comes from by just writing out the first few derivatives in general form:

$$\frac{d}{dx}\left(x^k \right) = k x^{k-1}$$

$$\frac{d^2}{dx^2}\left(x^k \right) = k(k-1) x^{k-2}$$

...

$$\frac{d^a}{dx^a}\left(x^k \right) = k(k-1)...(k-a+1) x^{k-a}$$

This is true for any k. Then if k was an integer we'd notice that:

$$k(k-1)...(k-a+1) = \frac{k(k-1)...(k-a+1)(k-a)...(2)(1)}{(k-a)...(2)(1)} = \frac{k!}{(k-a)!}$$

which gives the formula. When k is not an integer, you can't do this. But when k is 1/2, you can do something like it:

$$\frac{1}{2}\left(\frac{1}{2}-1\right) = \frac{1}{2}\left(-\frac{1}{2}\right) = - \frac{1}{4}$$

$$\frac{1}{2}\left(\frac{1}{2}-1\right) \left(\frac{1}{2}-2\right) = \frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) = \frac{3}{8}$$

Can you see where this is going, and come up with a formula for k(k-1)...(k-a+1) when k=1/2?

 

[RocketSurgery]

Thanks for the quick reply. This problem seems somewhat complicated for 1st yr calc. I don't know why my book has so many problems like this so early (This is the first introduction of Implicit Differentiation and Higher Order Derivatives in the book) and the book considers them easy problems. Your method makes the problem seem a little easier though. I'll follow that line of thinking and see if I can solve the problem.

 

[RocketSurgery]

I tried finding a formula for when k=1/2 but I'm still pretty confused.

I'm pretty sure this is wrong but this is what I came up with:

.5(.5-a-1)x^(.5-a)= (-.5a-.5)x^(.5-a)Plus I'm not sure how to apply this to my problem since the x isn't by itself and you would normally need to use chain rule.

 

[HallsofIvy]

?? Where did the "a" in the exponent come from? In your problem you just have an exponent of 1/2.

You know that the derivative of x^k is k x^k-1. Further, you know, by the chain rule, that the derivative of (x+5)^k is k(x+5)^k times the derivative of x+5, which is 1. That is, the derivative of (x+5)^1/2 is (1/2)(x+5)^1/2-1= (1/2)(x+5)^-1/2. The second derivative is the derivative of that: (1/2)[(-1/2)(x+5)^-1/2-1= (1/2)(-1/2)(x+5)^-3/2. Calculate a few more and see if you can find a pattern.

 

[RocketSurgery]

lol I have no idea where the "a" came from this whole thing is new to me I just started learning how to do these types of problems. I'm having a hard time figuring out the pattern. Even looking at the answer in the book I have absolutely no idea how you could find the answer without guess and check. The solution they list is:

y^(n)={[(-1)^(n+1)][1x3x5...(2n-3)]/[2^n]}{x+5}^[-(2n-1)/2]I'm trying to understand how to approach these types of problems but I am utterly confused.Edit: Actually now I remember why I have an "a" it stands for the "a"th derivative but I'm sure I'm still not in the right direction.

 

[RocketSurgery]

Ok I was working on this a little more today and came up with this:


Where "a" is the "a"th derivative

[(.5-{a-1})(1/2-{a-2})...(.5-0)(x^{.5-a})]

Btw thanks for the insights guys! The reason I was having trouble was because I thought I could use some kind of universal nth derivative formula but I realized it would be different depending on the function.

 

[HallsofIvy]

If f(x)= (x+5)^(1/2) then, by the power rule
f"(x)= (1/2)(x+5)^(-1/2)
f"(x)= (1/2)(-1/2)(x+5)^-3(3/2)
f"'(x)= (1/2)(-1/2)(-3/2) x^(-3/2)
f^IV(x)= (1/2)(-1/2)(-3/2)x^(-5/2)
f^V(x)= (1/2)(-1/2)(-3/2)x^(-5/2)
...
Looks to me like f⁽ⁿ⁾= (-1)^(n-1)(n-1)!/2^n-2 x^-n/2.

 

[RocketSurgery]

You're going to need to throw some brackets in there so I can understand your answer. I appreciate the help but from the pattern all I can deduce is what I put down and I've checked it and works for all the derivatives I tried however I am aware it is probably able to be condensed using factorials. I'm done with this problem though in hopes that I will understand it more after I take calc this fall until then I'm going to keep working on trying to come up with some proofs for some calc rules to get sum practice on those..