Find y' for y = (3^x)/(2^x)

Find $y'$ for $y = \dfrac{3^{x}}{2^{x}}$.

Step 1: Simplify using the laws of exponents

Because the two exponents are equal, the quotient is itself a single exponential: $y = \frac{3^{x}}{2^{x}} = \left(\frac{3}{2}\right)^{\!x}$.

This avoids the quotient rule entirely.

Step 2: Apply the exponential rule

Using $\bigl(a^{x}\bigr)' = a^{x}\ln a$ with $a = \dfrac{3}{2}$: $y' = \left(\frac{3}{2}\right)^{\!x}\ln\frac{3}{2}$.

Step 3: Rewrite in the original notation

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

Step 4: Verification

The quotient rule gives the same answer, at the cost of more work: \[y' = \frac{3^{x}\ln 3\cdot 2^{x} - 3^{x}\cdot 2^{x}\ln 2}{\bigl(2^{x}\bigr)^{2}} = \frac{3^{x}\bigl(\ln 3 - \ln 2\bigr)}{2^{x}} \;\;\checkmark\]

\[x = 0:\quad y = 1,\ y' = \ln\frac{3}{2} \approx 0.4055; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 0.4055 \;\;\checkmark\]