Use logarithmic differentiation to find the derivative of the function. $y=\sqrt[3]{\frac{x^{8}+1}{x-1}}$
A) $\frac{7 x^{8}-8 x^{7}+1}{3\left(x^{8}+1\right)^{2 / 3}(x-1)^{4 / 3}}$
B) $\frac{7 x^{8}-8 x^{7}-1}{3\left(x^{8}+1\right)^{2 / 3}(x-1)^{4 / 3}}$
C) $\frac{7 x^{8}-8 x^{7}+1}{3\left(x^{8}+1\right)(x-1)}$
D) $\frac{7 x^{8}-8 x^{7}-1}{3\left(x^{8}+1\right)(x-1)}$

Question

Quizplus · Accepted Answer

We can use logarithmic differentiation to find the derivative. First, we take the natural logarithm of both sides:

$$\ln(y) = \ln\left(\sqrt[3]{\frac{x^{8}+1}{x-1}}\right) = \frac{1}{3}\ln\left(\frac{x^{8}+1}{x-1}\right)$$

Next, we differentiate implicitly with respect to $x$ and simplify:

$$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{3} \cdot \frac{1}{\frac{x^{8}+1}{x-1}} \cdot \frac{d}{dx}\left(\frac{x^{8}+1}{x-1}\right)$$

Using the quotient rule and simplifying:

$$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{3} \cdot \frac{(8x^7)(x-1) - (x^8+1)}{(x-1)^2(x^8+1)}$$

Solving for $\frac{dy}{dx}$, we get:

$$\frac{dy}{dx} = \frac{y}{3} \cdot\frac{(8x^7)(x-1) - (x^8+1)}{(x-1)^2(x^8+1)}$$

Now we can substitute $y=\sqrt[3]{\frac{x^{8}+1}{x-1}}$:

$$\frac{dy}{dx} = \frac{\sqrt[3]{\frac{x^{8}+1}{x-1}}}{3} \cdot\frac{(8x^7)(x-1) - (x^8+1)}{(x-1)^2(x^8+1)}$$

Simplifying the numerator, we get:

$$\frac{dy}{dx} = \frac{7x^8 - 8x^7 - 1}{3(x^8+1)^{2/3}(x-1)^{4/3}}$$

Therefore, the answer is B, $\frac{7x^8 - 8x^7 - 1}{3(x^8+1)^{2/3}(x-1)^{4/3}}$.

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Use Logarithmic Differentiation to Find the Derivative of the Function $y=\sqrt[3]{\frac{x^{8}+1}{x-1}}$