Find y' for y = (ln 3 ·sin x+ cos x)/(3^x)

Find $y'$ for $y = \dfrac{\ln 3\cdot\sin x+\cos x}{3^{x}}$.

Step 1: Identify the parts and apply the quotient rule

For $y = \dfrac{u}{v}$ recall $y' = \dfrac{u'v - uv'}{v^{2}}$. Since $\ln 3$ is a constant coefficient: $u = \ln 3\cdot\sin x+\cos x,\quad u' = \ln 3\cdot\cos x - \sin x$.

\[ v = 3^{x},\quad v' = 3^{x}\ln 3. \]

Step 2: Substitute

\[ y' = \frac{\bigl(\ln 3\cdot\cos x - \sin x\bigr)3^{x} - \bigl(\ln 3\cdot\sin x+\cos x\bigr)3^{x}\ln 3}{3^{2x}}. \]

Step 3: Cancel one factor of $3^{x}$

\[ y' = \frac{\bigl(\ln 3\cdot\cos x-\sin x\bigr) - \ln 3\bigl(\ln 3\cdot\sin x+\cos x\bigr)}{3^{x}}. \]

Step 4: Expand and simplify the numerator

The two $\ln 3\cdot\cos x$ terms cancel, leaving only sine terms: \[\ln 3\cos x - \sin x - \ln^{2} 3\sin x - \ln 3\cos x = -\sin x - \ln^{2} 3\sin x = -\bigl(1+\ln^{2} 3\bigr)\sin x.\]

Step 5: Verification

\[x = 0:\quad u = 1,\ u' = \ln 3,\ v = 1,\ v' = \ln 3 \;\Longrightarrow\; y'(0) = \frac{\ln 3 - 1\cdot\ln 3}{1} = 0\]

\[\text{Formula:}\quad y'(0) = -\frac{\bigl(1+\ln^{2}3\bigr)\sin 0}{1} = 0 \;\;\checkmark\]

\[x = \frac{\pi}{2}:\quad y' = -\frac{1+\ln^{2}3}{3^{\pi/2}} \approx -\frac{2.2069}{5.6151} \approx -0.3931 \;\;\checkmark\]