Find the derivative of y = (x - 1)^x.

Step-by-step solution

1. 1

   Recognise why the standard rules fail. The power rule $\frac{d}{dx}x^n=nx^{n-1}$ needs a constant exponent, and the exponential rule $\frac{d}{dx}a^x=a^x\ln a$ needs a constant base. Here both the base $x-1$ and the exponent $x$ vary, so neither rule applies and mixing them gives nonsense. The fix is logarithmic differentiation.

2. 2

   State the domain first. For $(x-1)^x$ to be defined as a real power with a varying exponent we need a positive base:

   $x-1>0\qquad\Longleftrightarrow\qquad x>1$

   Everything below is valid on $(1,\infty)$, which is also exactly where $\ln(x-1)$ makes sense.

3. 3

   Take natural logarithms of both sides. The logarithm converts the exponent into a coefficient, which is the whole point:

   $\ln y=\ln\bigl((x-1)^x\bigr)=x\ln(x-1)$

   The right-hand side is now an ordinary product of two familiar functions.

4. 4

   Differentiate both sides with respect to $x$. On the left, $y$ is a function of $x$, so the chain rule gives $\frac{1}{y}\,y'$ — this is where implicit differentiation enters. On the right, use the product rule:

   $\frac{d}{dx}\bigl(x\ln(x-1)\bigr)=1\cdot\ln(x-1)+x\cdot\frac{1}{x-1}$

   so

   $\frac{y'}{y}=\ln(x-1)+\frac{x}{x-1}$

5. 5

   Solve for $y'$ and substitute the original function back. Multiplying by $y$ and replacing $y=(x-1)^x$:

   $\boxed{\,y'=(x-1)^x\left(\ln(x-1)+\frac{x}{x-1}\right)\,}$

   The answer must contain the original expression as a factor — that is typical of logarithmic differentiation, and a missing $(x-1)^x$ out front is the usual error.

6. 6

   Check numerically. At $x=3$ the formula gives $2^3\left(\ln 2+\tfrac{3}{2}\right)=8(0.6931+1.5)\approx 17.545$. A central difference quotient with $h=10^{-6}$ on $(x-1)^x$ at $x=3$ returns $17.5452$, so the derivative is right. The same template works for any $f(x)^{g(x)}$: its derivative is $f^g\left(g'\ln f+\frac{gf'}{f}\right)$.