## Calculus (3rd Edition)

$$y =\frac{1}{2}(x-4)+\ln 16$$
Given $$f(x)=\ln \left(x^{2}\right) \quad x=4$$ Since at $x=4$, $f(x) = \ln 16$ and $$f'(x)= \frac{2x}{x^2}= \frac{2}{x}$$ Then $m= f'(x)\bigg|_{x=4}=\dfrac{1}{2}$ Hence, the tangent line is given by \begin{align*} \frac{y-y_1}{x- x_1}&=m\\ \frac{y-\ln 16 }{x- 4}&=\dfrac{1}{2} \\ y&=\frac{1}{2}(x-4)+\ln 16 \end{align*}