Chapter 4 - Calculating the Derivative - 4.5 Derivatives of Logarithmic Functions - 4.5 Exercises - Page 241: 52

Use the fact that $\frac{d \ln x}{dx}=\frac{1}{x}$ , the change-of base theorem for logarithms, is $$ \log_{a}x=\frac{\ln x}{\ln a} $$ Find the derivative of each side, by the quotient rule , we have: $$ \begin{aligned} \frac{d \log _{a} x}{d x} &=\frac{\ln a \cdot \frac{d \ln x}{d x}-\ln x \cdot \frac{d \ln a}{d x}}{(\ln a)^{2}} \\ &=\frac{\ln a \cdot \frac{1}{x}-\ln x \cdot 0}{(\ln a)^{2}} \\ &=\frac{\frac{1}{x}}{\ln a} \\ &=\frac{1}{x \ln a}. \end{aligned} $$

Use the fact that $\frac{d \ln x}{dx}=\frac{1}{x}$ , the change-of base theorem for logarithms, is $$ \log_{a}x=\frac{\ln x}{\ln a} $$ Find the derivative of each side, by the quotient rule , we have: $$ \begin{aligned} \frac{d \log _{a} x}{d x} &=\frac{\ln a \cdot \frac{d \ln x}{d x}-\ln x \cdot \frac{d \ln a}{d x}}{(\ln a)^{2}} \\ &=\frac{\ln a \cdot \frac{1}{x}-\ln x \cdot 0}{(\ln a)^{2}} \\ &=\frac{\frac{1}{x}}{\ln a} \\ &=\frac{1}{x \ln a}. \end{aligned} $$