differentiate f and find the domain of f. (enter the domain in interval notation.) f(x) = ln(x^2 - 14x)…

Explanation:

Step1: Differentiate using chain - rule

Let $u = x^{2}-14x$, then $f(x)=\ln(u)$. The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$, and the derivative of $u = x^{2}-14x$ with respect to $x$ is $u'=2x - 14$. By the chain - rule $\frac{d}{dx}f(x)=\frac{d}{du}\ln(u)\cdot\frac{du}{dx}$, so $f'(x)=\frac{2x - 14}{x^{2}-14x}=\frac{2(x - 7)}{x(x - 14)}$.

Step2: Find the domain of $f(x)$

For $y = \ln(x^{2}-14x)$, the argument of the logarithm must be positive, i.e., $x^{2}-14x>0$. Factoring gives $x(x - 14)>0$. The solutions of the equation $x(x - 14)=0$ are $x = 0$ and $x = 14$. Using a sign - chart or testing intervals, the inequality is satisfied when $x<0$ or $x>14$. So the domain is $(-\infty,0)\cup(14,\infty)$.

Step3: For the tangent line problem

First, differentiate $y=\ln(x^{2}-6x + 1)$. Let $u=x^{2}-6x + 1$, then $\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dx}=2x-6$. By the chain - rule, $\frac{dy}{dx}=\frac{2x - 6}{x^{2}-6x + 1}$. Evaluate the derivative at $x = 6$: $\frac{dy}{dx}\big|{x = 6}=\frac{2(6)-6}{6^{2}-6(6)+1}=\frac{12 - 6}{1}=6$. The equation of a line in point - slope form is $y - y{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(6,0)$ and $m = 6$. So $y-0=6(x - 6)$, which simplifies to $y=6x-36$.

Answer:

derivative: $f'(x)=\frac{2(x - 7)}{x(x - 14)}$ domain: $(-\infty,0)\cup(14,\infty)$ tangent line: $y = 6x-36$