Question

If $y={\log }_{e}x(x-2{)}^2$ for $x\ne 0,2$ then $y^{\text{'}}\left(3\right)=?$

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{4}{3}$
• D. none of these

Answer 1

The correct option is D

none of these

Explanation for the correct option.

Step 1: Simplify

Given that, $y={\log }_{e}x(x-2{)}^2$.

Using the property of logarithmic i.e; ${\log }_{a}b=\frac{\log b}{\log a}$.

$\begin{array}{rcl}y& =& \frac{\log x{\left(x-2\right)}^2}{\log e}\\ & =& \log x{\left(x-2\right)}^2;\left[\log e=1\right]\\ & =& \log x+\log {\left(x-2\right)}^2;[\log ab=\log a+\log b]\\ & =& \log x+2\log \left(x-2\right)\end{array}$

Step 2: Differentiate

Now differentiate $y$ with respect to $x$ we get:

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{x}+\frac{2}{\left(x-2\right)}\\ \frac{dy}{dx}{|}_{x=3}& =& \frac{1}{3}+\frac{2}{\left(3-2\right)}\\ & =& \frac{1}{3}+2\\ & =& \frac{7}{3}\end{array}$

Hence, option D is correct.