How do I find the maclaurin series of $f(x)=\ln { (1+4x) } $

Question

Find the Maclaurin series for $f(x)$ using the definition of a Maclaurin series. [Assume that $f$ has a power series expansion. Do not show that $R_n(x) → 0$]

$f(x)=\ln { (1+4x) } $

What I did:

$$f(x)=\ln { (1+4x) } \\ f'(x)=\frac { 4 }{ 1+4x } =4(1+4x)^{ -1 }\\ f''(x)=-16(1+4x)^{ -2 }\\ f'''(x)=128(1+4x)^{ -3 }\\ f^{ (4) }(x)=-1536(1+4x)^{ -4 }$$

I could not figure out, or see what the $n$th derivative is.

I also got that:

$$f(0)=\ln { (1) } =0\\ f'(0)=4\\ f''(0)=-16\\ f'''(0)=128\\ f^{ (4) }(0)=-1536\\ $$

I would appreciate some help/guidance that would help me arrive at the correct solution on my own. I know the Maclaurin series for $f(x)=\ln { (1+x) } $, but I think the $4x$ is throwing me off here.

Answer 1

Observe that $f^{(n)}(x) = \dfrac{(-1)^{n+1}4^n(n-1)!}{(1+4x)^n}$.

Answer 2

So if you know the general form for the $n$th derivative of $f(x)=\ln(1+x)$, what do we have to include to make the $n$th derivative $f(x)=\ln(1+4x)$?

Maybe try comparing the $1$st derivative of $\ln(1+x)$ and the $1$st derivative of $\ln(1+4x)$, then the second derivative of $\ln(1+x)$ with the second derivative of $\ln(1+4x)$, and so on...