# Thread: nth derivative of a function

1. ## nth derivative of a function

I need to find that $\displaystyle n^{th}$ derivative of the function $\displaystyle f(x)=\frac{1}{(2-x)^2}$. I'm not really sure how to proceed. I wrote down a few derivatives, but I can't find a pattern. Is their a general method or formula for $\displaystyle n^{th}$ derivatives? Thanks.

I need to find that $\displaystyle n^{th}$ derivative of the function $\displaystyle f(x)=\frac{1}{(2-x)^2}$. I'm not really sure how to proceed. I wrote down a few derivatives, but I can't find a pattern. Is their a general method or formula for $\displaystyle n^{th}$ derivatives? Thanks.
$\displaystyle f^{\prime}(x)=\frac{2}{(2-x)^3}$

$\displaystyle f^{\prime\prime}(x)=\frac{6}{(2-x)^4}$

$\displaystyle f^{\prime\prime\prime}(x)=\frac{24}{(2-x)^5}$

...

So it follows that $\displaystyle f^{(n)}(x)=\frac{(n+1)!}{(2-x)^{n+2}}$.

3. Originally Posted by skeeter
Chris forgot a (-1) factor ... easy fix.

$\displaystyle f^{\prime}(x)=\frac{-2}{(2-x)^3}$

$\displaystyle f^{\prime\prime}(x)=\frac{6}{(2-x)^4}$

$\displaystyle f^{\prime\prime\prime}(x)=\frac{-24}{(2-x)^5}$

...

$\displaystyle f^{(n)}(x)=\frac{(-1)^n(n+1)!}{(2-x)^{n+2}}$.
Don't forget to chain the denominator! $\displaystyle \frac{\,d}{\,dx}(2-x)=-1$! Hence, all the negatives are cancelled!

So

$\displaystyle f^{\prime}(x)=\frac{-2}{(2-x)^3}(-1)=\frac{2}{(2-x)^3}$

$\displaystyle f^{\prime}(x)=\frac{-6}{(2-x)^{4}}(-1)=\frac{6}{(2-x)^4}$

etc...

4. my error ... sorry.

5. Originally Posted by skeeter
my error ... sorry.
However, I see your point since $\displaystyle (2-x)^2=(x-2)^2$...

If we rewrote the function as $\displaystyle f(x)=\frac{1}{(x-2)^2}$, then you would be correct....

6. Originally Posted by Chris L T521
However, I see your point since $\displaystyle (2-x)^2=(x-2)^2$...

If we rewrote the function as $\displaystyle f(x)=\frac{1}{(x-2)^2}$, then you would be correct....
Ok, that's why I was having difficulty find a pattern. I didn't notice $\displaystyle (2-x)^2=(x-2)^2$.

Ok, that's why I was having difficulty find a pattern. I didn't notice $\displaystyle (2-x)^2=(x-2)^2$.
It would also make this easier to recognize that $\displaystyle f(x)= \frac{1}{(2-x)^2}= \frac{1}{(x-2)^2}= (x- 2)^{-2}$.
$\displaystyle f'(x)= -2(x-2)^{-3}$
$\displaystyle f"(x)= 6(x-2)^{-4}= (-1)^2(2+1)!)(x-2)^{-(2+2)}$
$\displaystyle f"'(x)= -12(x-2)^{-5}= (-1)^3(3+1)!)(x-2)^{-(3+2)}$
$\displaystyle f""(x)= 60(x-2)^{-6}= (-1)^4(4+1)!(x-2)^{-(4+2)}$