use logarithmic differentiation to find the derivative of y with respect to the given independent…

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln\left[\sqrt[5]{(x^{5}+4)(x - 4)^{5}}\right]=\frac{1}{5}\ln\left[(x^{5}+4)(x - 4)^{5}\right]$ Using the property $\ln(ab)=\ln a+\ln b$, we get $\ln y=\frac{1}{5}\left[\ln(x^{5}+4)+5\ln(x - 4)\right]=\frac{1}{5}\ln(x^{5}+4)+\ln(x - 4)$.

Step2: Differentiate both sides with respect to $x$

The derivative of $\ln y$ with respect to $x$ is $\frac{1}{y}y'$ by the chain - rule. The derivative of $\frac{1}{5}\ln(x^{5}+4)$ using the chain - rule: Let $u = x^{5}+4$, then $\frac{d}{dx}\left[\frac{1}{5}\ln u\right]=\frac{1}{5}\cdot\frac{1}{u}\cdot\frac{du}{dx}=\frac{1}{5}\cdot\frac{1}{x^{5}+4}\cdot5x^{4}=\frac{x^{4}}{x^{5}+4}$. The derivative of $\ln(x - 4)$ is $\frac{1}{x - 4}$. So, $\frac{1}{y}y'=\frac{x^{4}}{x^{5}+4}+\frac{1}{x - 4}$.

Step3: Solve for $y'$

Multiply both sides by $y$: $y'=y\left(\frac{x^{4}}{x^{5}+4}+\frac{1}{x - 4}\right)$. Since $y=\sqrt[5]{(x^{5}+4)(x - 4)^{5}}$, we have $y'=\sqrt[5]{(x^{5}+4)(x - 4)^{5}}\left(\frac{x^{4}}{x^{5}+4}+\frac{1}{x - 4}\right)$.

Answer:

$y'=\sqrt[5]{(x^{5}+4)(x - 4)^{5}}\left(\frac{x^{4}}{x^{5}+4}+\frac{1}{x - 4}\right)$