Questions: Use logarithmic differentiation to find the derivative of the function. y = sqrt((x-3)/(x^6+1)) y' =

Transcript text: Use logarithmic differentiation to find the derivative of the function. \[ \begin{array}{l} y=\sqrt{\frac{x-3}{x^{6}+1}} \\ y^{\prime}=\square \end{array} \]

Solution

Solution Steps

Step 1: Take the natural logarithm on both sides

\[\ln(y) = \ln\left(\frac{\left(x - 3\right)^{0.5}}{x^{6} + 1}\right)\]

Step 2: Use properties of logarithms to simplify

\[\ln(y) = 0.5 \log{\left(x - 3 \right)} - \log{\left(x^{6} + 1 \right)}\]

Step 3: Differentiate both sides with respect to \(x\)

\[\frac{d}{dx}(\ln(y)) = - \frac{6 x^{5}}{x^{6} + 1} + \frac{0.5}{x - 3}\]

Step 4: Solve for \(\frac{{dy}}{{dx}}\) by multiplying both sides by \(y\)

\[\frac{dy}{dx} = \frac{0.577 \cdot \left(0.5 x^{6} + 6 x^{5} \cdot \left(3 - x\right) + 0.5\right)}{\left(0.333 x - 1\right)^{0.5} \left(x^{6} + 1\right)^{2}}\]

Final Answer:

\[\frac{dy}{dx} = \frac{0.577 \cdot \left(0.5 x^{6} + 6 x^{5} \cdot \left(3 - x\right) + 0.5\right)}{\left(0.333 x - 1\right)^{0.5} \left(x^{6} + 1\right)^{2}}\]

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