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

Explanation:

Step1: Take natural - log of both sides

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

Step2: Differentiate both sides with respect to $x$

$\frac{1}{y}\cdot\frac{dy}{dx}=\frac{1}{5}\cdot\frac{5x^{4}}{x^{5}+4}+\frac{1}{x - 4}$ $\frac{1}{y}\cdot\frac{dy}{dx}=\frac{x^{4}}{x^{5}+4}+\frac{1}{x - 4}$

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

$\frac{dy}{dx}=y\left(\frac{x^{4}}{x^{5}+4}+\frac{1}{x - 4}\right)$ Substitute $y=(x^{5}+4)^{\frac{1}{5}}(x - 4)$ back in: $\frac{dy}{dx}=(x^{5}+4)^{\frac{1}{5}}(x - 4)\left(\frac{x^{4}}{x^{5}+4}+\frac{1}{x - 4}\right)$ $=(x^{5}+4)^{\frac{1}{5}}\cdot\frac{x^{4}(x - 4)}{x^{5}+4}+(x^{5}+4)^{\frac{1}{5}}$ $=\frac{x^{4}(x - 4)}{(x^{5}+4)^{\frac{4}{5}}}+(x^{5}+4)^{\frac{1}{5}}$

Answer:

$\frac{x^{4}(x - 4)}{(x^{5}+4)^{\frac{4}{5}}}+(x^{5}+4)^{\frac{1}{5}}$