find the equation of the line tangent to the graph of f(x) = (ln x)^3 at x = 9.\ny = \n(type your answer in…

Explanation:

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

Using the chain - rule, if $y = u^{3}$ and $u=\ln x$, then $\frac{dy}{du}=3u^{2}$ and $\frac{du}{dx}=\frac{1}{x}$. So $f^\prime(x)=\frac{3(\ln x)^{2}}{x}$.

Step2: Find the slope of the tangent line at $x = 9$

Substitute $x = 9$ into $f^\prime(x)$. $m=f^\prime(9)=\frac{3(\ln 9)^{2}}{9}=\frac{(\ln 9)^{2}}{3}$.

Step3: Find the y - coordinate of the point on the curve at $x = 9$

Substitute $x = 9$ into $f(x)$. $y_0=f(9)=(\ln 9)^{3}$.

Step4: Use the point - slope form $y - y_0=m(x - x_0)$

Here $x_0 = 9$, $y_0=(\ln 9)^{3}$, and $m=\frac{(\ln 9)^{2}}{3}$. $y-(\ln 9)^{3}=\frac{(\ln 9)^{2}}{3}(x - 9)$ $y=\frac{(\ln 9)^{2}}{3}x-3(\ln 9)^{2}+(\ln 9)^{3}$ $\ln 9\approx2.1972$, $(\ln 9)^{2}\approx4.8377$, $(\ln 9)^{3}\approx10.6349$ $y=\frac{4.8377}{3}x-3\times4.8377 + 10.6349$ $y\approx1.61x-14.5131+10.6349$ $y\approx1.61x - 3.88$

Answer:

$y\approx1.61x - 3.88$