$n$th derivative of $x^3(x-2x^{1/2})^2$

Question

I don't know how to find the n th derivative of this equation

because there are negative integer so I don't how to use factorial on it.

I Try(?):

$\left(f(x)g(x)\right)^{(n)}=\sum_{k=0}^n\binom nk f^{(k)}(x)\cdot g^{(n-k)}(x)$

Let $f(x) = x^4\ and \ g(x) = (\sqrt(x)-2)^2 )$

$( x^4(\sqrt(x)-2)^2)^{(n)}=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{4!}{(4-n)!}x^{(4-n)}\cdot \frac{(\sqrt(x)-2)^2\cdot2!}{(x-n+k)!}$

Is this?

Start by taking the derivatives of $x^{1/2}$ and see if you can detect pattern. — Maesumi, Sep 28 '13 at 16:29
@Maesumi : I detect the pattern but couldn't find out how I can use factorial on it. — IndyZa, Sep 28 '13 at 17:15

score 3 · Answer 1 · answered Sep 28 '13 at 16:11

3

Hints:

$$\begin{align*}(1)&\;\;\;x^3\left(x-2\sqrt x\right)^2=x^4\left(\sqrt x-2\right)^2\\{}\\ (2)&\;\;\text{If}\;\;f(x)\,,\,g(x)\;\;\text{are two differentiable functions $\;n\;$ times, then}\\&\;\;\; \left(f(x)g(x)\right)^{(n)}=\sum_{k=0}^n\binom nk f^{(k)}(x)\cdot g^{(n-k)}(x)\end{align*}$$

Number (2) is called Leibniz Rule.

answered Sep 28 '13 at 16:11

DonAntonio

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What is the ∑k=0n(nk) ? I don't know how to calculate this.. :/ – IndyZa Sep 28 '13 at 16:18
I can't understand what you wrote. Are you saying you don't know the summatory symbol $;\sum;$ , or maybe the symbol for the binomial coefficient $;\binom nk;$ ? – DonAntonio Sep 28 '13 at 16:20
The (n k) (binomial coefficient). – IndyZa Sep 28 '13 at 16:22
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But you tagged your question as "calculus" and "derivatives": you should know this! Anyway, the algebraic definition is $$\binom nk:=\frac{n!}{k!(n-k)!}$$ If you want another definition, more geometrical and, imo, prettier, google "Pascal's Triangle". Very nice, elementary and beautiful stuff! There's even a third, combinatorial definition...perhaps in aset theory or something like that. – DonAntonio Sep 28 '13 at 16:25

score 3 · Answer 2 · edited Apr 13 '17 at 12:21

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Hint: It is easier to expand it

$$ f(x) = {x}^{5}-4\,{x}^{9/2}+4\,{x}^{4}. $$

Now, it is easy to find the $n$th derivative.

Added; Here is an example

$$ (x^5)^{(n)}= \frac{5!}{(5-n)!} x^{5-n}$$

Note: You can use the general formula

$$ \frac{d^n}{dx^n} x^m = \frac{\Gamma(m+1)}{\Gamma(m-n+1 )} x^{m-n}=\frac{m!}{(m-n )!} x^{m-n}\,. $$

edited Apr 13 '17 at 12:21

Community

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answered Sep 28 '13 at 16:14

Mhenni Benghorbal

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I don't know how to write it on factorial if the derivative is more than 4 (the power become -) – IndyZa Sep 28 '13 at 16:18
The example can only handle up to n=5 right? since there are no negative factorial... – IndyZa Sep 28 '13 at 16:25
@IndyZa: I recommand you to find few derivatives of the function $x^m$ to see what's going on. – Mhenni Benghorbal Sep 28 '13 at 16:28
I see... How about the m that is not integer.. It can not fits with the factorial right? – IndyZa Sep 28 '13 at 16:40
@IndyZa: Yes. Check the generalization of the factorial which is the gamma function. – Mhenni Benghorbal Sep 28 '13 at 16:45
I really don't understand the gamma function on how it works...so its Γ(7/2+1) supposed if m = 7/2 then what just put 9/2!? – IndyZa Sep 28 '13 at 16:50

Maesumi · Accepted Answer · 2013-09-28T21:26:34.370

Take the first 6 derivatives "by hand" until the derivatives of the polynomial portion disappear. Now for the rest I give a hint and you modify it to apply it to your problem. Suppose you want to take the derivatives of $y=x^{1/2}$.

you get

$y^{(0)}=x^{1/2}$

$y^{(1)}=1/2 x^{-1/2}$

$y^{(2)}=(1/2)*(-1/2) x^{-3/2}$

$y^{(3)}=(1/2)*(-1/2)*(-3/2) x^{-5/2}$

$y^{(4)}=(1/2)*(-1/2)*(-3/2)*(-5/2) x^{-7/2}$

$\cdots$

$y^{(n)}=(1/2)*(-1/2)*(-3/2)*(-5/2)*\cdots *(1/2-(n-1)) x^{(1/2)-n}$

Now to understand this derivative we need several pieces. (Assume we are looking at derivatives of order $1$ or higher.)

(1) For derivatives of even order product is negative, for odd order it is positive. So this sign can be reproduced by a $(-1)^{n-1}$.

(2) The exponent of $x$ decreases from $-1/2$ to $1/2-n$, i.e. for the $n$-th order it is $-(2n-1)/2$.

(3) The coefficient, aside from the sign, can be written as $(1/2)*(1/2)*(3/2)*(5/2)*\cdots *(2n-3)/2$, where the numerator is the product of all odd numbers from $1$ to $(2n-3)$ and the denominator is $n$ copies of $2$. So the denominator is $2^n$. Now to clean and write the numerator with factorials you can apply the following trick: Multiply and divide by all the numbers that are missing from making a complete factorial in the numerator.

So

$1*3*5*\cdots(2n-3)= \displaystyle{1*2*3*4*5*\cdots*(2n-3)*(2n-2) \over 2*4*6*\cdots*(2n-2)}$

Now notice that you can factor a $2$ from every term in the denominator and write it as $2^{n-1}(n-1)!$. And the numerator is nicely $(2n-2)!$. Putting it all together for $n>0$ we have

$({\sqrt x})^{(n)}= \displaystyle{ (-1)^{n-1} {{(2n-2)!} \over {2^{2n-1}(n-1)!}} x^{-(2n-1)/2}}$

Now to your problem!

To start: $f(x)=x^5+4x^4-4x^{9/2}$. So we just differentiate.

$f'=5x^4+4*4x^3-4*(9/2)x^{7/2}$

$f''=5*4x^3+4*4*3*x^2-4*(9/2)*(7/2)x^{5/2}$

$f'''=5*4*3x^2+4*4*3*2*x-4*(9/2)*(7/2)*(5/2)x^{3/2}$

$f''''=5*4*3*2x+4*4*3*2-4*(9/2)*(7/2)*(5/2)*(3/2)x^{1/2}$

or $f^{iv}= 5!x+4*4!-4*(9!/4!*2^8)x^{1/2}$

$f^{v}=5!-4*(9!/4!*2^9)x^{-1/2}$

For $n>1$ we have $f^{(n+4)}=-4*(9!/4!*2^8) \displaystyle{ (-1)^{n-1} {{(2n-2)!} \over {2^{2n-1}(n-1)!}} x^{-(2n-1)/2}} $

$n$th derivative of $x^3(x-2x^{1/2})^2$

3 Answers3