differentiate.\ny = \\frac{\\ln x}{x^{3}}\n\\frac{dy}{dx}=\\square\n(use parentheses to clearly denote the…

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \ln x$ and $v=x^{3}$.

Step2: Find $u^\prime$ and $v^\prime$

The derivative of $u=\ln x$ is $u^\prime=\frac{1}{x}$ (by the derivative formula for $\ln x$), and the derivative of $v = x^{3}$ is $v^\prime=3x^{2}$ (using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$).

Step3: Apply the quotient - rule

$y^\prime=\frac{\frac{1}{x}\cdot x^{3}-\ln x\cdot3x^{2}}{(x^{3})^{2}}$. Simplify the numerator: $\frac{1}{x}\cdot x^{3}=x^{2}$ and the denominator $(x^{3})^{2}=x^{6}$. So $y^\prime=\frac{x^{2}-3x^{2}\ln x}{x^{6}}$. Factor out $x^{2}$ from the numerator: $y^\prime=\frac{x^{2}(1 - 3\ln x)}{x^{6}}$. Cancel out $x^{2}$ in the numerator and denominator: $y^\prime=\frac{1 - 3\ln x}{x^{4}}$.

Answer:

$\frac{1 - 3\ln x}{x^{4}}$