Use logarithmic differentiation to find the derivative of y.
-$$y = x ( x - 7 ) ( x - 1 )$$
A) $x ( x - 7 ) ( x - 1 ) ( \ln x + \ln ( x - 7 ) + \ln ( x - 1 ) )$
B) $\frac { 1 } { x } + \frac { 1 } { x - 7 } + \frac { 1 } { x - 1 }$
C) $x ( x - 7 ) ( x - 1 ) \left( \frac { 1 } { x } + \frac { 1 } { x - 7 } + \frac { 1 } { x - 1 } \right)$
D) 1

Question

Quizplus · Accepted Answer

1. Take the natural logarithm of both sides:

$$\ln y = \ln (x(x-7)(x-1))$$

2. Apply the rules of logarithms:

$$\ln y = \ln x + \ln (x-7) + \ln (x-1)$$

3. Differentiate both sides with respect to x using the chain rule and product rule on the right-hand side:

$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \frac{1}{x-7} + \frac{1}{x-1}$$

$$\frac{dy}{dx} = y \left( \frac{1}{x} + \frac{1}{x-7} + \frac{1}{x-1} \right)$$

4. Substitute y with its original expression:

$$\frac{dy}{dx} = x(x-7)(x-1) \left( \frac{1}{x} + \frac{1}{x-7} + \frac{1}{x-1} \right)$$

Therefore, the correct option is C: $$\boxed{C. } x(x-7)(x-1) \left( \frac{1}{x} + \frac{1}{x-7} + \frac{1}{x-1} \right)$$

Use Logarithmic Differentiation to Find the Derivative of Y y=x(x−7)(x−1)y = x ( x - 7 ) ( x - 1 )y=x(x−7)(x−1)

Use Logarithmic Differentiation to Find the Derivative of Y $y = x ( x - 7 ) ( x - 1 )$