Let $$ f(x) = (x+1)log(x+1), \text{$x\in]-1,\infty[$} $$ where log is the natural log.
a) Show that $$f^{(n)}(x) = \frac{(-1)^n(n-2)!}{(x+1)^{n-1}} \ \text{for $n\ge 2$ and $x\in]-1,\infty[ $}$$
The Taylor polynomial for $f$ of degree $3$ at the point $0$ is given by: $$(p_3f)(x) = x+\frac{1}{2}x^2-\frac{1}{6}x^3$$
b) Show that for $n\ge2$ and $x\ge0$ the remainder is given by the formula: $$ R_nf(x) = \frac{(-1)^{n+1}}{n}\int_0^x\frac{(x-t)^n}{(t+1)^n}dt $$
a) I know that the n'th derivative of $g(x) =log(x+1)$ is given by: $$g^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{(1+x)^n}$$ So the n'th derivative of $f(x)$ is given by: $$f^{(n)} = (x+1)\cdot\frac{(-1)^{n-1}(n-1)!}{(1+x)^n} = \frac{(-1)^{n-1}(n-1)!}{(1+x)^{n-1}}$$
I am not quite sure how to go on from here with a)
b) I now that the Remainder can be expressed as: $$R_nf(x) = f(x) -T_n(x)$$ and $$R_nf(x) = \frac{1}{n!}\int_0^x f^{(n+1)}(t)(x-t)^ndt $$ I know that I somehow have to use the second formula to show the desired in c), but I am not sure how to do it properly.