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Chapter 3: Problem 59

Calculate the derivative of the following functions. $$y=\frac{1}{\log _{4} x}$$

Short Answer

Expert verified
Question: Find the derivative of the following function: $$y=\frac{1}{\log _{4} x}$$ Solution: The derivative of the given function is: $$\frac{dy}{dx} = \frac{\ln 4}{x (\ln x)^2}$$

Step by step solution

01

Rewrite the Function

Rewrite the given function using properties of logarithms: $$y=\frac{1}{\log _{4} x} = \frac{1}{\frac{\ln x}{\ln 4}} = \frac{\ln 4}{\ln x}.$$
02

Differentiate the Function using Chain Rule

To find the derivative, we apply the chain rule: $$\frac{dy}{dx} = \frac{d(\ln 4 \cdot (\ln x)^{-1})}{dx}.$$ First, we differentiate the outer function, treating \((\ln x)^{-1}\) as a separate function: $$\frac{d(\ln 4 \cdot (\ln x)^{-1})}{d(\ln x)^{-1}} = \ln 4 \cdot (-1) \cdot (\ln x)^{-2}.$$ Next, we differentiate the inner function: $$\frac{d(\ln x)^{-1}}{dx} = \frac{d(\ln x)^{-1}}{d\ln x} \cdot \frac{d\ln x}{dx} = (-1) \cdot (\ln x)^{-2} \cdot \frac{1}{x}.$$ Now applying the chain rule and combining the two differentiation results: $$\frac{dy}{dx} = (\ln 4 \cdot (-1) \cdot (\ln x)^{-2}) \cdot (-1) \cdot (\ln x)^{-2} \cdot \frac{1}{x}$$ $$\frac{dy}{dx} = \frac{\ln 4}{x (\ln x)^2}$$
03

Final Answer

The derivative of the given function is: $$\frac{dy}{dx} = \frac{\ln 4}{x (\ln x)^2}$$

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