use logarithmic differentiation to find the derivative of y with respect to x.\ny = \\sqrt{\\frac{(x +…

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln\sqrt{\frac{(x + 3)^{8}}{(3x-1)^{12}}}=\frac{1}{2}\ln\frac{(x + 3)^{8}}{(3x-1)^{12}}=\frac{1}{2}(8\ln(x + 3)-12\ln(3x - 1))=4\ln(x + 3)-6\ln(3x - 1)$

Step2: Differentiate both sides with respect to x

$\frac{1}{y}\frac{dy}{dx}=4\cdot\frac{1}{x + 3}-6\cdot\frac{3}{3x - 1}$

Step3: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=y\left(\frac{4}{x + 3}-\frac{18}{3x - 1}\right)$ Substitute $y = \sqrt{\frac{(x + 3)^{8}}{(3x-1)^{12}}}$ back in: $\frac{dy}{dx}=\sqrt{\frac{(x + 3)^{8}}{(3x-1)^{12}}}\left(\frac{4}{x + 3}-\frac{18}{3x - 1}\right)$

Answer:

$\sqrt{\frac{(x + 3)^{8}}{(3x-1)^{12}}}\left(\frac{4}{x + 3}-\frac{18}{3x - 1}\right)$