Find the n -th derivative of y = x^(n-1)ln x

Find the $n$-th derivative of $y = x^{\,n-1}\ln x$.

Step 1: Small cases first

\[ n = 1:\ y = \ln x,\ y' = \frac{1}{x} = \frac{0!}{x}; \qquad n = 2:\ y = x\ln x,\ y'' = \frac{1}{x} = \frac{1!}{x}; \]

\[ n = 3:\ y = x^{2}\ln x,\ y''' = \frac{2}{x} = \frac{2!}{x}. \]

The answer to be proved is therefore $y^{(n)} = \dfrac{\bigl(n-1\bigr)!}{x}$.

Step 2: Set up the Leibniz formula

\[ y^{(n)} = \sum_{k=0}^{n}\binom{n}{k}\bigl(x^{\,n-1}\bigr)^{(k)}\bigl(\ln x\bigr)^{(n-k)}. \tag{1} \]

The two ingredients are

\[ \bigl(x^{\,n-1}\bigr)^{(k)} = \frac{\bigl(n-1\bigr)!}{\bigl(n-1-k\bigr)!}\,x^{\,n-1-k}\ \ \bigl(k\le n-1\bigr), \qquad \bigl(\ln x\bigr)^{(j)} = \frac{\bigl(-1\bigr)^{j-1}\bigl(j-1\bigr)!}{x^{j}}\ \ \bigl(j\ge1\bigr). \]

Since $x^{\,n-1}$ is a polynomial of degree $n-1$, its $n$-th derivative is zero, so the term $k = n$ drops out and the sum runs only to $n-1$. In particular every surviving term has $j = n-k\ge1$.

Step 3: Every term has the same shape

For $0\le k\le n-1$, with $j = n-k$: \[\binom{n}{k}\cdot\frac{\bigl(n-1\bigr)!}{\bigl(n-1-k\bigr)!}x^{\,n-1-k}\cdot\frac{\bigl(-1\bigr)^{\,n-k-1}\bigl(n-k-1\bigr)!}{x^{\,n-k}}.\]

Two simplifications occur at once: the factorials $\bigl(n-1-k\bigr)!$ and $\bigl(n-k-1\bigr)!$ are the same number and cancel, and the powers of $x$ combine to $x^{-1}$: $= \binom{n}{k}\bigl(n-1\bigr)!\,\bigl(-1\bigr)^{\,n-k-1}\cdot\frac1x$.

Step 4: Sum the binomial coefficients

\[ y^{(n)} = \frac{\bigl(n-1\bigr)!}{x}\sum_{k=0}^{n-1}\binom{n}{k}\bigl(-1\bigr)^{\,n-k-1}. \tag{2} \]

The binomial theorem gives $\displaystyle\sum_{k=0}^{n}\binom{n}{k}\bigl(-1\bigr)^{\,n-k} = \bigl(1-1\bigr)^{n} = 0$, so removing the last term, \[\sum_{k=0}^{n-1}\binom{n}{k}\bigl(-1\bigr)^{\,n-k} = -\binom{n}{n} = -1 \;\Longrightarrow\; \sum_{k=0}^{n-1}\binom{n}{k}\bigl(-1\bigr)^{\,n-k-1} = 1.\]

Step 5: Verification

\[n = 4:\ y = x^{3}\ln x,\ y' = 3x^{2}\ln x+x^{2},\ y'' = 6x\ln x+5x,\ y''' = 6\ln x+11,\ y^{\mathrm{IV}} = \frac{6}{x} = \frac{3!}{x} \;\;\checkmark\]

\[n = 5:\ y^{(5)} = \frac{4!}{x} = \frac{24}{x} \;\;\checkmark\]

\[\text{The cases } n = 2 \text{ and } n = 3 \text{ are Problems 275 and 280} \;\;\checkmark\]