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Find the $n^{th}$ derivative of $$f(x) = e^x\cdot x^n$$

If i am not wrong i have following

$1^{st}$ Derivative: $e^x\cdot n\cdot x^{n-1} + x^n\cdot e^x$

$2^\text{nd}$ Derivative: $e^x\cdot n\cdot (n-1)\cdot x^{n-2} + 2 \cdot e^x\cdot n\cdot x^{n-1} + x^n\cdot e^x$

$3^\text{rd}$ Derivative: $e^x\cdot n\cdot (n-1)\cdot (n-2)\cdot x^{n-3} + 3 \cdot e^x\cdot n \cdot (n-1)\cdot x^{n-2} + 4 \cdot e^x \cdot n \cdot x^{n-1} + 2 \cdot x^n\cdot e^x$

From here how do I calculate the $n^{th}$ derivative?

Thanks. :)

rndflas
  • 955

2 Answers2

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Note that $$\frac{d^m}{dx^m}(e^xx^n)=\sum_{p=0}^m{\binom{m}{p}\frac{d^{m-p}}{dx^{m-p}}(e^x)\frac{d^p}{dx^p}(x^n)}$$ This is the generalized Leiniz rule. This is more succinctly written as $$\frac{d^m}{dx^m}(e^xx^n)=\sum_{p=0}^m{\frac{n!}{(n-p)!}\binom{m}{p}e^xx^{n-p}}$$

Matt Samuel
  • 58,164
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You have derived the solution for yourself, but simply did not complete the last step:

$$ f^{(k)}(x) = \frac{n! e^x\cdot x^{n-k}}{(n-k)!} + f^{(k-1)}(x) $$

therefore

$$ f^{(k-1)}(x) = \frac{n! e^x\cdot x^{n-k+1}}{(n-k+1)!} + f^{(k-2)}(x) $$

and so on

$$ f^{(k-2)}(x) = \frac{n! e^x\cdot x^{n-k+2}}{(n-k+2)!} + f^{(k-3)}(x) $$

...

$$ f^{(k-i)}(x) = \frac{n! e^x\cdot x^{n-k+i}}{(n-k+i)!} + f^{(k-1-i)}(x) $$

If you sum them up you get

$$ f^{(k)}(x) = e^x \sum\limits_{i=0}^{k} \frac{n!}{(n-k+i)!} \cdot x^{n-k+i} $$

bkosztin
  • 300