hw 14 - derivatives of logs section 2.9: problem 10 (1 point)\nfind an equation of the line tangent to the…

Explanation:

Step1: Find the derivative of the function

The derivative of $y = 6\ln(x)$ using the formula $\frac{d}{dx}\ln(x)=\frac{1}{x}$ is $y'=\frac{6}{x}$.

Step2: Find the slope of the tangent line at $x = 6$

Substitute $x = 6$ into $y'$. So $m=y'(6)=\frac{6}{6}=1$.

Step3: Find the y - coordinate when $x = 6$

Substitute $x = 6$ into $y = 6\ln(x)$. So $y=6\ln(6)$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(6,6\ln(6))$ and $m = 1$. $y-6\ln(6)=1\times(x - 6)$ $y=x-6 + 6\ln(6)$

Answer:

$y=x-6 + 6\ln(6)$