Find the derivative of \( y \) with respect to \( r \) \( y=\log _{4} r \cdot \log _{6} r \) \( \frac{d y}{d r}= \)

To find the derivative of \( y = \log_{4} r \cdot \log_{6} r \), we can use the product rule. First, let's rewrite the logarithms in terms of natural logarithms for easier differentiation: \[ y = \frac{\ln r}{\ln 4} \cdot \frac{\ln r}{\ln 6} \] Now, apply the product rule: \[ \frac{d y}{d r} = \frac{d}{d r} \left( \frac{\ln r}{\ln 4} \right) \cdot \frac{\ln r}{\ln 6} + \frac{\ln r}{\ln 4} \cdot \frac{d}{d r} \left( \frac{\ln r}{\ln 6} \right) \] Differentiating \( \ln r \) gives us \( \frac{1}{r} \): \[ \frac{d}{d r} \left( \frac{\ln r}{\ln 4} \right) = \frac{1}{r \ln 4} \] \[ \frac{d}{d r} \left( \frac{\ln r}{\ln 6} \right) = \frac{1}{r \ln 6} \] Substituting back in: \[ \frac{d y}{d r} = \frac{1}{r \ln 4} \cdot \frac{\ln r}{\ln 6} + \frac{\ln r}{\ln 4} \cdot \frac{1}{r \ln 6} \] Combining these two terms, we can factor out the common parts: \[ \frac{d y}{d r} = \frac{1}{r} \left( \frac{\ln r}{\ln 4 \cdot \ln 6} + \frac{\ln r}{\ln 4 \cdot \ln 6} \right) \] \[ \frac{d y}{d r} = \frac{2 \ln r}{r \ln 4 \cdot \ln 6} \] Thus, the final answer is: \[ \frac{d y}{d r} = \frac{2 \ln r}{r \ln 4 \cdot \ln 6} \]