Question

Consider the function f(x) = x ln(1 + x).

Find an expression for the k-th derivative of f(x) when k  2.

Answer 1

Let's compute the first two derivatives to see if there's a pattern:

f '(x) = (x/(1+x)) +ln(1+x)

f ''(x) = ((1+x) - x)/(1+x)^2 + 1/(1+x)

= (2+x)/(1+x)^2, when k=2

Answer 2

If you need the pattern for k in general let me know. But that is the answer for k = 2

Answer 3

I can see what you did there though I can't seem to figure out the expression for k in general. If you could show me that it would be great.

Answer 4

Sorry didn't see in my question above it was meant to be when \[k \ge2\]

Answer 5

Ok hold on

Answer 6

\[f(x)=x\cdot\ln{(1+x)}\]\[f^{(k)}(x)=\sum\limits_{j=0}^k{k\choose j}(x)^{(j)}\cdot(\ln{(1+x)})^{(k-j)}=\]\[=1\cdot(x)^{(0)}\cdot(\ln{(1+x)})^{(k)}+k\cdot(x)^{(1)}\cdot(\ln{(1+x)})^{(k-1)}=\]\[=x\cdot(-1)^{k-1}\frac{(k-1)!}{(1+x)^k}+k\cdot(-1)^{k-2}\frac{(k-2)!}{(1+x)^{k-1}}=\]\[=(-1)^{k-2}\frac{(k-2)!}{(1+x)^{k-1}}\cdot\left(-(k-1)\frac{x}{1+x}+k\right)=\]\[=(-1)^{k-2}\frac{(k-2)!}{(1+x)^{k-1}}\cdot\frac{-kx+x+k+kx}{1+x}=\]\[=(-1)^{k-2}(k-2)!\frac{x+k}{(1+x)^k}\]