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

$\ln y=\frac{1}{2}\left[\ln(x + 2)^{12}-\ln(2x - 5)^{9}\right]$ Using the property $\ln\frac{a}{b}=\ln a-\ln b$ and $\ln a^{n}=n\ln a$.

Step2: Simplify the right - hand side

$\ln y=\frac{1}{2}(12\ln(x + 2)-9\ln(2x - 5))=6\ln(x + 2)-\frac{9}{2}\ln(2x - 5)$

Step3: Differentiate both sides with respect to $x$

$\frac{1}{y}\frac{dy}{dx}=6\cdot\frac{1}{x + 2}-\frac{9}{2}\cdot\frac{2}{2x - 5}$ Using the chain - rule $\frac{d}{dx}\ln u=\frac{1}{u}\frac{du}{dx}$.

Step4: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=y\left(\frac{6}{x + 2}-\frac{9}{2x - 5}\right)$ Substitute $y = \sqrt{\frac{(x + 2)^{12}}{(2x - 5)^{9}}}$ back in: $\frac{dy}{dx}=\sqrt{\frac{(x + 2)^{12}}{(2x - 5)^{9}}}\left(\frac{6}{x + 2}-\frac{9}{2x - 5}\right)$

Answer:

$\sqrt{\frac{(x + 2)^{12}}{(2x - 5)^{9}}}\left(\frac{6}{x + 2}-\frac{9}{2x - 5}\right)$