Question

find the equation of the line tangent to the graph of (y = 4ln(x)) at (x = 2). tangent line: (y=)

Explanation:

Step1: Find the derivative of the function

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

Step2: Evaluate the derivative at $x = 2$

Substitute $x = 2$ into $y'$. So $y'(2)=\frac{4}{2}=2$. This is the slope $m$ of the tangent - line.

Step3: Find the $y$ - coordinate of the point of tangency

Substitute $x = 2$ into $y = 4\ln(x)$. So $y=4\ln(2)$. The point of tangency is $(2,4\ln(2))$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(2,4\ln(2))$ and $m = 2$.
$y-4\ln(2)=2(x - 2)$.
Expand to get $y-4\ln(2)=2x-4$.
Then $y=2x+4\ln(2)-4$.

Answer:

$y = 2x+4\ln(2)-4$