# Thread: find the n'th differentiation of these..

1. ## find the n'th differentiation of these..

1=<k=<n
find $\displaystyle f^{(n)} (x)$ of:

the first :
$\displaystyle f(x) = |x|^{n + 1}$
the second is:
$\displaystyle f(x) = |\sin x|^{n + 1}$

2. The first function k-derivate is $\displaystyle f^{k}(x)=\begin{cases}\displaystyle\frac{(n+1)!}{( n+1-k)!}x^{n+1-k},x>0\\ \\ (-1)^{n+1}\displaystyle\frac{(n+1)!}{(n+1-k)!}x^{n+1-k},x<0\\\end{cases}$.

The second involves a lots of fun because you'll have to do the same with a product !

3. what is the general way you found this result??

what is the derivative of absolute value function??

4. how you find the n'th derivative
did you use macloren series formula
??

5. Originally Posted by Abu-Khalil
The first function k-derivate is $\displaystyle f^{k}(x)=\begin{cases}\displaystyle\frac{(n+1)!}{( n+1-k)!}x^{n+1-k},x>0\\ \\ (-1)^{n+1}\displaystyle\frac{(n+1)!}{(n+1-k)!}x^{n+1-k},x<0\\\end{cases}$.

The second involves a lots of fun because you'll have to do the same with a product !
how you got this solution??

6. when i use leibnitz formula

for the second question:
$\displaystyle (|\sin x|^{n}||\sin x|)^{(n)}=\sum_{k=0}^{n}C^k_n (|\sin x|^{n})^{(n-k)}|sinx|^{(k)}\\$

what to do next??

7. Originally Posted by transgalactic
how you got this solution??
Just observing what happened every time i derivated the function and tried to make a rule.

8. what about the solution i proposed??

9. Originally Posted by transgalactic
its similar to
when i use leibnitz formula
in this case g(x)=1
$\displaystyle (fg)^{(n)}=\sum_{k=0}^{n}C^k_n |x|^{{n + 1}^{(n-k)}g^{(k)}}$
so by this solution
i just need to input f(x) there

The first function k-derivate is $\displaystyle f^{k}(x)=\begin{cases}\displaystyle\frac{(n+1)!}{( n+1-k)!}x^{n+1-k},x>0\\ \\ (-1)^{n+1}\displaystyle\frac{(n+1)!}{(n+1-k)!}x^{n+1-k},x<0\\\end{cases}$.