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I'm trying to find the n-order derivative of a function.

Function: $f(x) = x^3.lnx$

1.Derivative: $f'(x) = x^2.(3lnx+1)$ 2.Derivative: $f''(x) = x.(6lnx+5)$ 3.Derivative: $f'''(x) = (6lnx+11)$ 4.Derivative: $f^{(4)}(x) = 6x^{-1}$
5.Derivative: $f^{(5)}(x) = -6x^{-2}$
6.Derivative: $f^{(6)}(x) = 12x^{-3}$
7.Derivative: $f^{(7)}(x) = -36x^{-4}$ 8.Derivative: $f^{(8)}(x) = 144x^{-5}$

How can I write the general derivative of this function over n only thing I can find is for $x$

n.Derivative: $f^{(n)} = x^{n-3}..?$

Just guide me, I believe I will figure out the rest myself.

Baran Kanat
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  • Write $f^{(n)} = c_n x^{n-3}$ and try to find a recursion formula for the $c_n$. – Martin R Dec 05 '22 at 09:27
  • @MartinR Thanks, but I'm not very knowledgeable about recursive functions, please do you have more details or a resource you can provide? – Baran Kanat Dec 05 '22 at 09:30
  • $f^{(4)}(x) = 6x^{-1}$, $f^{(5)}(x) = 6(-1)x^{-2}$, $f^{(6)}(x) = 6(-1)(-2)x^{-3}$ , $f^{(7)}(x) = 6(-1)(-2)(-3)x^{-4}$ – can you recognize a pattern? – Martin R Dec 05 '22 at 09:32
  • @MartinR Thank you very much my friend, now I understand The 4th derivative and after will fit as you said. $f^{(n)} = x^{3-n}.(n-4)!$ What about the first three derivatives? – Baran Kanat Dec 05 '22 at 09:40

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