# Computing the nth derivative using the binomial theorem.

Let n be an element of the set of natural numbers

Let $$F(x)=x^2(1+x)^n$$ and write $$F^n$$ for the nth derivative of the function $$F$$.

Compute $$F^n$$ by applying the Binomial Theorem to $$(1+x)^n$$.

I don't understand the step where I need to find the derivative of these terms:

= $$x^2 + {n \choose 1}x^3 + {n \choose 2} x^4 + .... + {n \choose n-1} x^{n+1} + x^{n+2}$$

Even though I do understand how to show that for the first 3 terms the nth derivative is 0, but for the last two I have no clue??

Really appreciate any hints!!

The $$n$$-th derivative is $$0$$ for many more than just the first three terms: in fact, for all but the last three.

You have a typo, the last term is $$x^{n+2}$$.

The last three terms are $${n\choose{n-2}} x^n+{n\choose{n-1}} x^{n+1}+x^{n+2}$$

Can you work it out from here? Some factorials will appear when you take the $$n$$-th derivative, hence you might want to express the $$n\choose k$$ in terms of factorials as well, to see if something cancels out, and at the very least to get a nice-looking formula.

• Would you explain why just for the last three terms? – Shervan Oct 31 at 11:16

Write it in decreasing powers of $$x$$ and use subscript $$n$$: $$F_n(x)=x^2(1+x)^n=x^2(x+1)^n=\\ \color{red}{x^{n+2}}+\color{green}{nx^{n+1}}+\color{blue}{\frac{n(n-1)}{2}x^{n}}+\frac{n(n-1)(n-2)}{6}x^{n-1}+\cdots+nx^3+x^2$$ Note that for $$f(x)=x^n$$: $$f^{(0)}(x)=f(x)=x^n\\ f^{(k)}(x)=\begin{cases}n(n-1)(n-2)\cdots(n-k+1)x^{n-k},k\le n\\ 0,k>n\end{cases}$$ Hence: $$F_0(x)=x^2 \Rightarrow F_0^{(0)}(x)=F_0(x)=x^2;\\ F_n^{(n)}(x)=\color{red}{(n+2)(n+1)n\cdots3\cdot x^2}+\color{green}{n\cdot (n+1)n\cdots1\cdot x}+\\ \color{blue}{\frac{n(n-1)}{2}\cdot n(n-1)\cdots 1}=\\ \frac12(n+2)(n+1)\cdot n!\cdot x^2+n(n+1)\cdot n!\cdot x+\frac{n(n-1)}{2}\cdot n!=\\ n!\left(\frac12(n+2)(n+1)x^2+n(n+1)x+\frac{n(n-1)}{2}\right).$$ Verify: $$F_1(x)=x^2(x+1)=x^3+x^2 \Rightarrow F_1^{(1)}(x)=1!(3x^2+2x)=3x^2+2x;\\ F_2(x)=x^2(x+1)^2=x^4+2x^3+x^2 \Rightarrow \\ F_2^{(2)}(x)=2!(6x^2+6x+1)=12x^2+12x+2.$$

Since $$\deg F=n+2$$, we should get a quadratic. Use the generalized product rule$$\frac{d^n}{dx^n}g(x)h(x)=\sum_{k=0}^n\binom{n}{k}g^{(k)}h^{(n-k)}.$$For $$g=x^2$$, only $$0\le k\le 2$$ contributes, giving$$x^2\frac{d^n}{dx^n}(1+x)^n+2nx\frac{d^{n-1}}{dx^{n-1}}(1+x)^n+n(n-1)\frac{d^{n-2}}{dx^{n-2}}(1+x)^n.$$Since $$\frac{d^{n-k}}{dx^{n-k}}(1+x)^n$$ is $$\frac{n!}{k!}(1+x)^k$$, the above simplifies to$$n!x^2+n!2nx(1+x)+\frac{n!}{2}n(n-1)(1+x)^2\\=n!\left(\frac12n(n-1)+n(n+1)x+\frac12(n+1)(n+2)x^2\right).$$

• sanity check: for $n=1$, your formula gives $x^2+2x$ while it must be $3x^2+2x$. – farruhota Oct 29 at 18:42
• @farruhota Thanks; fixed. – J.G. Oct 29 at 18:56