using logarithmic differentiation, find the derivative of the function y = - 3x^2 / (x - 7)^(1/4) provide…

Explanation:

Step1: Take natural - log of both sides

First, rewrite $y =-\frac{3x^{2}}{\sqrt[4]{x - 7}}$ as $y=-3x^{2}(x - 7)^{-\frac{1}{4}}$. Then $\ln y=\ln(-3x^{2}(x - 7)^{-\frac{1}{4}})=\ln3 + 2\ln x-\frac{1}{4}\ln(x - 7)$ (ignoring the negative sign for now as we will deal with it later when differentiating the original function).

Step2: Differentiate both sides with respect to $x$

The derivative of $\ln y$ with respect to $x$ is $\frac{y'}{y}$ by the chain - rule. The derivative of $\ln3$ is $0$ (since $\ln3$ is a constant), the derivative of $2\ln x$ is $\frac{2}{x}$, and the derivative of $-\frac{1}{4}\ln(x - 7)$ is $-\frac{1}{4(x - 7)}$. So, $\frac{y'}{y}=\frac{2}{x}-\frac{1}{4(x - 7)}$.

Step3: Solve for $y'$

Multiply both sides by $y=- \frac{3x^{2}}{\sqrt[4]{x - 7}}$: [ \begin{align*} y'&=y\left(\frac{2}{x}-\frac{1}{4(x - 7)}\right)\ &=-\frac{3x^{2}}{\sqrt[4]{x - 7}}\left(\frac{2}{x}-\frac{1}{4(x - 7)}\right)\ &=-\frac{3x^{2}}{\sqrt[4]{x - 7}}\times\frac{2}{x}+\frac{3x^{2}}{\sqrt[4]{x - 7}}\times\frac{1}{4(x - 7)}\ &=-\frac{6x}{\sqrt[4]{x - 7}}+\frac{3x^{2}}{4(x - 7)^{\frac{5}{4}}} \end{align*} ]

Answer:

$y'=-\frac{6x}{\sqrt[4]{x - 7}}+\frac{3x^{2}}{4(x - 7)^{\frac{5}{4}}}$