Find y' for y = u^v, where u = phi(x) > 0 and v = psi(x)

Find $y'$ for $y = u^{v}$, where $u = \varphi(x) \gt 0$ and $v = \psi(x)$ are differentiable.

Step 1: Take the natural logarithm

\[ \ln y = v\ln u. \]

Step 2: Differentiate both sides

The right-hand side is a product of two functions of $x$, and $\ln u$ needs the chain rule: $\frac{y'}{y} = v'\ln u + v\cdot\frac{u'}{u}$.

Step 3: Solve for $y'$

\[ y' = u^{v}\!\left(v'\ln u + \frac{v\,u'}{u}\right). \]

Step 4: Split into two recognisable pieces

Distributing $u^{v}$ over the bracket and using $\dfrac{u^{v}}{u} = u^{v-1}$: \[y' = \underbrace{v\,u^{v-1}u'}_{\text{power-rule part}} + \underbrace{u^{v}\ln u\cdot v'}_{\text{exponential-rule part}}.\]

The first term is what one gets by treating $v$ as constant, the second by treating $u$ as constant. The derivative is the sum of the two — a two-variable chain rule in disguise.

Step 5: Verification — the three special cases

\[v = \text{const}:\quad v' = 0 \;\Longrightarrow\; y' = v\,u^{v-1}u', \text{ the power rule} \;\;\checkmark\]

\[u = \text{const}:\quad u' = 0 \;\Longrightarrow\; y' = u^{v}\ln u\cdot v', \text{ the exponential rule} \;\;\checkmark\]

\[u = v = x:\quad y' = x\cdot x^{x-1}\cdot 1 + x^{x}\ln x\cdot 1 = x^{x}\bigl(1+\ln x\bigr), \text{ Problem 175} \;\;\checkmark\]