$\text { Use logarithmic differentiation to find the derivative of } y = \frac { x ^ { 36 } \sqrt { 29 x - 4 } } { ( x - 1 ) ^ { 14 } } \text {. }$
A) $\frac { d y } { d x } = \frac { x ^ { 36 } \sqrt { 29 x - 4 } } { ( x - 1 ) ^ { 14 } }$
B) $\frac { d y } { d x } = \frac { x ^ { 36 } \sqrt { 29 x - 4 } } { ( x - 1 ) ^ { 14 } } \left( \frac { 36 } { x } + \frac { 29 } { 2 ( 29 x - 4 ) } - \frac { 14 } { x - 1 } \right)$
C) $\frac { d y } { d x } = \frac { x ^ { 36 } \sqrt { 29 x - 4 } } { ( x - 1 ) ^ { 14 } } \left( \frac { 36 } { x } - \frac { 29 } { 2 ( 29 x - 4 ) } - \frac { 14 } { x - 1 } \right)$
D) $\frac { d y } { d x } = \frac { 36 x \sqrt { 29 x - 4 } } { 14 ( x - 1 ) ^ { 13 } }$
E) $\frac { d y } { d x } = \frac { x ^ { 36 } \sqrt { 29 x - 4 } } { ( x - 1 ) ^ { 14 } } \left( \frac { 36 } { x } - \frac { 29 } { 2 ( 29 x - 4 ) } + \frac { 14 } { x - 1 } \right)$

Question

$\text { Use logarithmic differentiation to find the derivative of } y = \frac { x ^ { 36 } \sqrt { 29 x - 4 } } { ( x - 1 ) ^ { 14 } } \text {. }$ 
A) $\frac { d y } { d x } = \frac { x ^ { 36 } \sqrt { 29 x - 4 } } { ( x - 1 ) ^ { 14 } }$
B) $\frac { d y } { d x } = \frac { x ^ { 36 } \sqrt { 29 x - 4 } } { ( x - 1 ) ^ { 14 } } \left( \frac { 36 } { x } + \frac { 29 } { 2 ( 29 x - 4 ) } - \frac { 14 } { x - 1 } \right)$
C) $\frac { d y } { d x } = \frac { x ^ { 36 } \sqrt { 29 x - 4 } } { ( x - 1 ) ^ { 14 } } \left( \frac { 36 } { x } - \frac { 29 } { 2 ( 29 x - 4 ) } - \frac { 14 } { x - 1 } \right)$
D) $\frac { d y } { d x } = \frac { 36 x \sqrt { 29 x - 4 } } { 14 ( x - 1 ) ^ { 13 } }$
E) $\frac { d y } { d x } = \frac { x ^ { 36 } \sqrt { 29 x - 4 } } { ( x - 1 ) ^ { 14 } } \left( \frac { 36 } { x } - \frac { 29 } { 2 ( 29 x - 4 ) } + \frac { 14 } { x - 1 } \right)$

Quizplus · Accepted Answer

Taking the natural logarithm of both sides, we get:

$$\ln y = \ln\left(\frac{x^{36}\sqrt{29x-4}}{(x-1)^{14}}\right)$$

Using the properties of logarithms, we can simplify the right-hand side as:

$$\ln y = \ln(x^{36}\sqrt{29x-4}) - \ln(x-1)^{14}$$

$$\ln y = \ln x^{36} + \ln (29x-4)^{\frac12} - 14\ln(x-1)$$

$$\ln y = 36\ln x + \frac12\ln(29x-4) - 14\ln(x-1)$$

Now we differentiate both sides of this equation with respect to $x$ using the chain rule and product rule as needed:

$$\frac{1}{y}\frac{dy}{dx} = 36\frac{1}{x} + \frac12\frac{1}{29x-4}\cdot29 - 14\frac{1}{x-1}$$

$$\frac{1}{y}\frac{dy}{dx}=\frac{36}{x}+\frac{29}{2(29x-4)}-\frac{14}{x-1}$$

Multiplying both sides by $y$, we get:

$$\frac{dy}{dx} = y \left(\frac{36}{x}+\frac{29}{2(29x-4)}-\frac{14}{x-1}\right)$$

Substituting the original expression for $y$:

$$\frac{dy}{dx} = \frac{x^{36}\sqrt{29x-4}}{(x-1)^{14}}\left(\frac{36}{x}+\frac{29}{2(29x-4)}-\frac{14}{x-1}\right)$$

Simplifying this expression, we get:

$$\frac{dy}{dx} = \frac{x^{36}\sqrt{29x-4}}{(x-1)^{14}}\left(\frac{36(x-1)(29x-4)+29x(x-1)-28(29x-4)}{2x(x-1)(29x-4)}\right)$$

$$\frac{dy}{dx} = \frac{x^{36}\sqrt{29x-4}}{(x-1)^{14}}\left(\frac{36(29x^2-62x+32)}{58x(x-1)(29x-4)}\right)$$

$$\frac{dy}{dx} = \frac{x^{36}\sqrt{29x-4}}{(x-1)^{14}}\left(\frac{18(29x-4-31x+16)}{29x(x-1)(29x-4)}\right)$$

$$\frac{dy}{dx} = \frac{x^{36}\sqrt{29x-4}}{(x-1)^{14}}\left(\frac{18}{x-1}-\frac{29}{2(29x-4)}+\frac{36}{x}\right)$$

Thus, the correct choice is (B).