Is this the correct inductive step to prove that the n-th derivative of ln(2x+1) is equal to my formula?

Question

I deduced that the n-th derivative is given through $f(x)^n=(-1)^{n+1}*\frac{2*2^{n-1}}{(2*x+1)^n}$. Is the correct inductive step $f(x)^{n+1}=(-1)^{n+2}*\frac{2*2^{n}}{(2*x+1)^{n+1}}$?

Answer 1

You should slow down and illustrate the steps explicitly. Currently you stated what you wanted to prove and I can't tell if you know why it is true.

Add in the steps $$f^{(n+1)}(x)=\frac{d}{dx}[f^{(n)}(x)]=\frac{d}{dx}\left[(-1)^{n+1} \frac{2\cdot 2^{n-1}}{(2x+1)^n}\right]$$

and go on and use the chain rule to verify.