Use logarithmic differentiation to find the derivative of y = (sin x)^(cos x).

Step-by-step solution

1. 1

   See why the ordinary rules fail. The power rule $\frac{d}{dx}u^n = nu^{n-1}u'$ needs a constant exponent, and the exponential rule $\frac{d}{dx}a^u = a^u \ln a \cdot u'$ needs a constant base. Here both base and exponent vary, so neither applies and logarithmic differentiation is the right tool.

2. 2

   Take the natural logarithm of both sides. The log turns the exponent into a factor:

   $\ln y = \cos x \cdot \ln(\sin x)$

   This is valid on intervals where $\sin x > 0$, for example $0 < x < \pi$, which is also where the original expression is defined for arbitrary real exponents.

3. 3

   Differentiate both sides implicitly. On the left, the chain rule gives $\dfrac{y'}{y}$. On the right, use the product rule on $\cos x$ and $\ln(\sin x)$:

   $\frac{y'}{y} = (-\sin x)\ln(\sin x) + \cos x \cdot \frac{\cos x}{\sin x}$

   The derivative of $\ln(\sin x)$ is $\dfrac{\cos x}{\sin x} = \cot x$ by the chain rule — this is the term most often dropped.

4. 4

   Tidy the right-hand side.

   $\frac{y'}{y} = \frac{\cos^2 x}{\sin x} - \sin x \ln(\sin x)$

5. 5

   Multiply back by $y$. Replace $y$ by the original expression — leaving the answer in terms of $y$ is incomplete:

   $y' = (\sin x)^{\cos x}\left(\frac{\cos^2 x}{\sin x} - \sin x \ln(\sin x)\right)$

   Equivalently, $y' = (\sin x)^{\cos x}\bigl(\cos x \cot x - \sin x \ln \sin x\bigr)$.

6. 6

   Verify at a convenient point. At $x = \tfrac{\pi}{4}$: $\sin x = \cos x = \tfrac{\sqrt2}{2} = 0.707107$, so $(\sin x)^{\cos x} = 0.707107^{0.707107} = 0.782100$. The bracket is $\dfrac{0.5}{0.707107} - 0.707107\ln(0.707107) = 0.707107 + 0.245095 = 0.952202$. Their product is $0.744716$. A symbolic derivative evaluated at $\tfrac{\pi}{4}$ gives $0.745221$ — the small gap is only the rounding in the hand figures above; carrying more digits ($0.7821020 \times 0.9528482$) reproduces $0.7452207$ ✓.