# Finding a general formula for the nth derivative of a partial fraction

Moved from technical math section, so missing the homework template
Summary:: Find a general formula for the nth derivative

Hi everyone!

How would I approach and answer a Q such as this I began by rewriting the expression in a different form, then used chain rule to each given term

I tried to work out the 1st, 2nd and 3rd derivative in order to help me spot a pattern, so that it would help me find a general formula for the nth derivative. But I can't seem to notice too much.

The only thing I see is that signs alternate as you differentiate again and again, and that it has a common factor of 1/5 to take out Any help would be appreciated, thanks!

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BvU
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2019 Award
If we take the 5 for granted (put it in front of brackets ##\ {1\over 5}(...)\ ## ) you have a generalization of the form $${d^n\over dx^n}= {1\over 5}\Bigl (\;A (x+1)^p + B (x-4)^q \;\Bigr )$$ and you have to find A, B, p and q. Not so difficult, especially p and q • Bolter
If we take the 5 for granted (put it in front of brackets ##\ {1\over 5}(...)\ ## ) you have a generalization of the form $${d^n\over dx^n}= {1\over 5}\Bigl (\;A (x+1)^p + B (x-4)^q \;\Bigr )$$ and you have to find A, B, p and q. Not so difficult, especially p and q Ok so I have come up with this now, following up from what you have shown I have multiplied my nth derivative my (-1)^n also, to account for the alternating sign changes
My p and q values must be -(n+1) from inspection

Although I still cannot what A and B values must be?
The only pattern that I can see now is that A & B get doubled then tripled then quadrupled then quintuple etc etc each time you differentiate the expression more and more

Does this involve something to do with factorials or I'm I mistaken here?

• BvU
BvU
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2019 Award
You are not mistaken !

You are not mistaken !
Ok yes, so I have used that in my nth derivative and got this You can probably see that I redone the differentiation for first, second and third derivative, but this time I did not simplify the coefficients and this is what helped me realised that I had to use the factorial notation I hope that is right nth derivative formula

PeroK
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Ok yes, so I have used that in my nth derivative and got this

View attachment 255915
You can probably see that I redone the differentiation for first, second and third derivative, but this time I did not simplify the coefficients and this is what helped me realised that I had to use the factorial notation I hope that is right nth derivative formula
One way to check the formula you have found is to use induction.

One way to check the formula you have found is to use induction.
I would but I haven't come across to using induction yet PeroK
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Gold Member
I would but I haven't come across to using induction yet There are two steps:

1) Show that your equation holds for n = 1 (possibly ##n=0## as well).

2) Show that if your equation holds for ##n = k## then it holds for ##n = k+1##.

Then you are done!

There are two steps:

1) Show that your equation holds for n = 1 (possibly ##n=0## as well).

2) Show that if your equation holds for ##n = k## then it holds for ##n = k+1##.

Then you are done!
Ahh so it's essentially a proof method then

I had given that a go and I do see that when n=1, it does give back the first order derivative that I found from before PeroK