Let $f(x)= a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. I am trying to find $f^n(x)$.

By applying the power rule $n$ times, I get this $$f^n(x)=a_{n}(n\cdot n)x^{n-n}+\cdots+ a_1$$ which I think can be simplified to $$f^n(x)=a_{n}n^2+\cdots+ a_1$$

However, I don't think I have the correct answer, as my exercise book is telling me the answer is. $$a_n\,n\cdot(n-1)\cdot\,\cdots\,\cdot 1$$

What did I do wrong? I'm under the impression I have done multiple mistakes.

  • 1
    $\begingroup$ Start with $\,\left(x^n\right)' = n x^{n-1}\,$, $\,\left(x^n\right)'' = n(n-1) x^{n-2}\,$, $\,\left(x^n\right)''' = n(n-1)(n-2) x^{n-3} \,\ldots\,$ $\endgroup$ – dxiv Jul 30 '18 at 0:34
  • $\begingroup$ Hint: $(x^n)' = n x^{n-1}$, but $(nx^{n-1})' = n(n-1) x^{n-2}$, and so on. Repeated application of the power rule give a falling factorial coefficient. $\endgroup$ – Xander Henderson Jul 30 '18 at 0:35
  • $\begingroup$ The exercise book should say the answer is $a_nn(n-1)...1$. $\endgroup$ – 高田航 Jul 30 '18 at 0:39
  • $\begingroup$ You are right, that is the answer in the book. I badly transcribed it, I modified my question. Thank you $\endgroup$ – Cedric Martens Jul 30 '18 at 0:41

Recall that the power rule for differentiation states:

Let $f(x)=x^n$ for some $n\in\mathbb{R}\backslash\lbrace 0\rbrace$. Then, $$\frac{d}{dx}f(x)=nx^{n-1}$$

Also recall that differentiation is linear in the sense that $$\frac{d}{dx}[f_1(x)+f_2(x)+\cdots+f_k(x)]=\frac{d}{dx}f_1(x)+\frac{d}{dx}f_2(x)+\cdots+\frac{d}{dx}f_k(x)$$ Now consider your function $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. Differentiating and applying linearity, we obtain $$\frac{d}{dx}f(x)=\frac{d}{dx}a_nx^n+\frac{d}{dx}a_{n-1}x^{n-1}+\cdots+\frac{d}{dx}a_1x+\frac{d}{dx}a_0$$ The constant term at the end becomes $0$, since the derivative of a constant is always $0$. The second to last term, $\frac{d}{dx}a_1x$ is simply $a_1$, since the derivative of a line is always its slope. As for the other terms, we simply apply the power rule. After completing the first round of differentiation, we should have something like this: $$f^1(x)=a_nnx^{n-1}+a_{n-1}x^{n-2}+\cdots+a_2x+a_1$$ If we keep repeating the differentiation process, you will see that there is always a constant term $a_m$ at the end which just becomes $0$. Since $f(x)$ has $n+1$ terms (each $a_ix^i$ from $i=0$ up to $i=n$), after $n$ rounds of differentiation, you should have eliminated all the terms except for the very first one: $a_nx^n$. Applying the power rule to just that term, we get: \begin{align} \frac{d}{dx}a_nx_n &=a_nnx^{n-1}\\ \frac{d}{dx}a_nnx^{n-1} & = a_nn(n-1)x^{n-2}\\ \frac{d}{dx}a_nn(n-1)x^{n-2}& =a_nn(n-1)(n-2)x^{n-3} \\ &\vdots \\ \frac{d}{dx}a_nn(n-1)(n-2)\cdots(3)x^2& =a_nn(n-1)(n-2)\cdots(3)(2)x \\ \frac{d}{dx}a_nn(n-1)(n-2)\cdots(3)(2)x&=a_nn(n-1)(n-2)\cdots(3)(2)(1) \\ & = a_n\cdot n! \end{align}

| cite | improve this answer | |

First, you should convince yourself that all the terms on the right after the first do not contribute because you eventually take the derivative of a constant, which is zero.

Then do it by hand for $n=3$. We want the third derivative of $a_3x^3$. The first derivative is $3a_3x^2$, the second is $3\cdot 2 a_3x$, the third is $3\cdot 2 \cdot 1a_3=3!a_3$

Now convince yourself that this is general and $f^n(x)=a_nn!$

Your proposed answer has two problems: the exponent decreases by $1$ each time you take a derivative so the second one should be $n-1$ and there are $n$ derivatives, not $2$. The numbers you bring down are multiplied, not added.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.