Question

1. - / 1 points use logarithmic differentiation to find the derivative of the function. y = x^7/x y(x) =

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln(x^{7/x})$. Using the property $\ln(a^b)=b\ln(a)$, we get $\ln y = \frac{7}{x}\ln x$.

Step2: Differentiate both sides with respect to $x$

The left - hand side: $\frac{d}{dx}(\ln y)=\frac{1}{y}y'$.
The right - hand side: Using the product rule $(uv)' = u'v+uv'$, where $u = \frac{7}{x}=7x^{-1}$ and $v=\ln x$.
$u'=- 7x^{-2}=-\frac{7}{x^{2}}$ and $v'=\frac{1}{x}$.
So, $\frac{d}{dx}(\frac{7}{x}\ln x)=-\frac{7}{x^{2}}\ln x+\frac{7}{x}\cdot\frac{1}{x}=-\frac{7\ln x}{x^{2}}+\frac{7}{x^{2}}=\frac{7(1 - \ln x)}{x^{2}}$.

Step3: Solve for $y'$

Since $\frac{1}{y}y'=\frac{7(1 - \ln x)}{x^{2}}$, and $y = x^{7/x}$, then $y'=y\cdot\frac{7(1 - \ln x)}{x^{2}}=x^{7/x}\cdot\frac{7(1 - \ln x)}{x^{2}}$.

Answer:

$x^{7/x}\cdot\frac{7(1 - \ln x)}{x^{2}}$