Find the n'th derivative of these

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

the first :
<br /> f(x) = |x|^{n + 1}<br />
the second is:
<br /> f(x) = |\sin x|^{n + 1} <br />
 
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What is the use of defining k?
Did you try calculating the first derivatives by hand and looking for a pattern?
 
what is the derivative of an absolute value function??
 
For you to solve an absolute value derivative you have to realize that there are 2 parts of the function: + and -. Therefore you take the derivative of both parts and set the regions where the derivative is true.

For example:
f(x)=|x|

The line y=-x exists from -infinity to 0 and the line y=x exists from 0 to +infinity. Then the derivative of f(x) is -1 on (-infinity, 0) and 1 on (0,infinity). Hope this helps!
 
in an ordinary function
the derivative from he left must equal the derivative from the right

so we can't do a derivative to |x|
because of that?? on one side i get 1 on the other i get -1
the same thing goes for the question i need to solve
??
 
Note that what you say is a pointwise statement.
Only for x = 0 you get different limits on both sides, which means that |x| is not differentiable at x = 0. This corresponds to the intuitive notion of differentiability, which |x| doesn't satisfy at x = 0 because it has a kink / sharp point there.
However, for x non-zero, the function is perfectly smooth and differentiable.

In fact, when you do analysis you start by defining what it means if f is differentiable at a point (see your other thread, where I gave the definition) and then for colloquial convenience define just "f is differentiable" to mean: "f is differentiable at all the points in the domain".
 
the makloren formula is
<br /> f(x)=\sum_{n}^{k}\frac{f^{(k)}}{k!}x^k+o(x^n)<br />

how to take the n'th derivative from here??

where to put the formula in and get the n'th derivative??

this function
is for approximating so in order for me to get to the n'th members approximation
first i need to do manually one by one n times derivative
so its not helping me
??
 
for the second question:
<br /> (|\sin x|^{n}||\sin x|)^{(n)}=\sum_{k=0}^{n}C^k_n (|\sin x|^{n})^{(n-k)}|sinx|^{(k)}\\<br />

what to do next??
 
Differntiating f(x)= |x|n+1 is easy. if x> 0 then f(x)= xn+1 what is the nth derivative of that? For example, if n= 3, (x4)'= 4x3, (x4)"= (4x3)'= 4(3)x2, (x4)'''= (4(3)x2)'= (4)(3)(2)x. Can you generalize that?

If x< 0, f(x)= -xn+1 whenever n+1 is odd.

The derivative of |x|n+1 at x= 0 is not defined when x= 0.
 
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