Question

find the equation of the line tangent to the graph of y = 4ln(x) at x = 3. tangent line: y = 4/3x - 4 + 4ln(3). attempt 2: 1 attempt remaining.

Explanation:

Step1: Find the derivative of the function

The derivative of $y = 4\ln(x)$ using the formula $\frac{d}{dx}\ln(x)=\frac{1}{x}$ is $y'=\frac{4}{x}$.

Step2: Evaluate the derivative at $x = 3$

Substitute $x = 3$ into $y'$. So $y'(3)=\frac{4}{3}$, which is the slope $m$ of the tangent - line.

Step3: Find the $y$ - value at $x = 3$

Substitute $x = 3$ into $y = 4\ln(x)$. So $y(3)=4\ln(3)$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(3,4\ln(3))$ and $m=\frac{4}{3}$.
$y-4\ln(3)=\frac{4}{3}(x - 3)$
$y-4\ln(3)=\frac{4}{3}x-4$
$y=\frac{4}{3}x-4 + 4\ln(3)$

Answer:

$y=\frac{4}{3}x-4 + 4\ln(3)$