find the equation of the line tangent to the graph of (f(x)=(ln x)^{2}) at (x = 6).\n(y=square) (type your…

Explanation:

Step1: Find the derivative of $f(x)$

Using the chain - rule. Let $u = \ln x$, then $y = u^{2}$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=2u$, and the derivative of $u$ with respect to $x$ is $\frac{du}{dx}=\frac{1}{x}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=2\ln x\cdot\frac{1}{x}=\frac{2\ln x}{x}$.

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

Substitute $x = 6$ into the derivative. $m=f^{\prime}(6)=\frac{2\ln 6}{6}=\frac{\ln 6}{3}$.

Step3: Find the $y$ - coordinate of the point on the curve at $x = 6$

Substitute $x = 6$ into $f(x)$. $y=f(6)=(\ln 6)^{2}$.

Step4: Use the point - slope form $y - y_{1}=m(x - x_{1})$ to find the equation of the tangent line

Here $x_{1}=6$, $y_{1}=(\ln 6)^{2}$, and $m = \frac{\ln 6}{3}$. [ \begin{align*} y-(\ln 6)^{2}&=\frac{\ln 6}{3}(x - 6)\ y-(\ln 6)^{2}&=\frac{\ln 6}{3}x-2\ln 6\ y&=\frac{\ln 6}{3}x-2\ln 6+(\ln 6)^{2} \end{align*} ] Calculate $\ln 6\approx1.79176$. [ \begin{align*} \frac{\ln 6}{3}&\approx\frac{1.79176}{3}\approx0.59725\ -2\ln 6&=-2\times1.79176=-3.58352\ (\ln 6)^{2}&=(1.79176)^{2}\approx3.2103 \end{align*} ] [ y\approx0.60x - 0.37 ]

Answer:

$y\approx0.60x - 0.37$