Power series derivative using binomial theorem

Question

Came across this question in my textbook:

$f(x) = (1+2x)^{10}$. Determine $f^{(5)}(0)$ using the binomial theorem.

If I am correct, the author of the book want me not to use the power rule. How else do I compute this?

Answer 1

Hint:

The binomial theorem says that $$(a+b)^n = \sum\limits_{k=0}^n {{n}\choose{k}} a^k b^{n-k}$$

Replace $a$ with $1$, $b$ with $2x$, and $n$ with $10$; you can then differentiated term by term in the summation.

Answer 2

You should get that $$(1+2x)^{10}=1024 x^{10} + 5120 x^9 + 11520 x^8 + 15360 x^7 + 13440 x^6 + 8064 x^5 + 3360 x^4 + 960 x^3 + 180 x^2 + 20 x + 1$$ and taking $5$ derivatives, you should get $$30965760 x^5 + 77414400 x^4 + 77414400 x^3 + 38707200 x^2 + 9676800 x + 967680$$ or $$967680 (2 x + 1)^5$$ where $f^{(5)}(0)$ is the coefficient of the original $x^5$ term undergoing power rule $5$ times to become $$967680=\binom{10}{5}\cdot2^5\cdot 5!$$