Question

question 5
if $f(x)=ln(5x + 1)^6$
find the slope of the tangent line to f(x) at x=-1

question 6
if $f(x)=ln(3x^{2}-5x - 2)$
find the slope of the tangent line to f(x) at x=2

Explanation:

Question 5

Step1: Simplify the function

Using the property $\ln a^b = b\ln a$, we can rewrite $f(x)=\ln(5x + 1)^6$ as $f(x)=6\ln(5x + 1)$.

Step2: Differentiate using the chain - rule

The derivative of $\ln(u)$ is $\frac{u'}{u}$. Let $u = 5x+1$, then $u'=5$. So $f'(x)=\frac{6\times5}{5x + 1}=\frac{30}{5x + 1}$.

Step3: Evaluate the derivative at $x=-1$

Substitute $x = - 1$ into $f'(x)$: $f'(-1)=\frac{30}{5\times(-1)+1}=\frac{30}{-4}=-\frac{15}{2}$.

Step1: Differentiate using the chain - rule

The derivative of $\ln(u)$ is $\frac{u'}{u}$. Let $u = 3x^2-5x - 2$, then $u'=6x - 5$. So $f'(x)=\frac{6x - 5}{3x^2-5x - 2}$.

Step2: Evaluate the derivative at $x = 2$

Substitute $x = 2$ into $f'(x)$: $f'(2)=\frac{6\times2-5}{3\times2^2-5\times2 - 2}=\frac{12 - 5}{12-10 - 2}=\frac{7}{0}$, which is undefined. But we can factor $u=3x^2-5x - 2=(3x + 1)(x - 2)$. Then $f'(x)=\frac{6x - 5}{(3x + 1)(x - 2)}$. After simplification, we use the quotient - rule or rewrite and differentiate. Another way is to use the chain - rule directly. $f'(x)=\frac{6x-5}{3x^2 - 5x - 2}$, substituting $x = 2$ gives $\frac{6\times2-5}{3\times2^2-5\times2 - 2}=\frac{7}{0}$. However, if we consider the limit as $x\to2$ of $f'(x)$. First, factor $3x^2-5x - 2=(3x + 1)(x - 2)$. Then $f'(x)=\frac{6x - 5}{(3x + 1)(x - 2)}$. Using L'Hopital's rule on the limit of the difference quotient (or by factoring and canceling common factors if possible). In this case, $f'(x)=\frac{6x-5}{3x^2-5x - 2}$, substituting $x = 2$:
We have $f'(x)=\frac{6x - 5}{3x^2-5x - 2}$, when $x = 2$, $f'(2)=\frac{6\times2-5}{3\times2^2-5\times2 - 2}=\frac{7}{0}$. But if we rewrite $f(x)=\ln(3x^2-5x - 2)$ and use the chain - rule: $f'(x)=\frac{6x-5}{3x^2-5x - 2}$. Since $3x^2-5x - 2=(3x + 1)(x - 2)$, we can also find the derivative as follows:
The derivative of $\ln(3x^2-5x - 2)$ is $\frac{6x - 5}{3x^2-5x - 2}$. Substitute $x = 2$:
$u = 3x^2-5x - 2$, $u'=6x - 5$. $f'(x)=\frac{u'}{u}=\frac{6x - 5}{3x^2-5x - 2}$. When $x = 2$, $3x^2-5x - 2=3\times4-10 - 2=0$. But if we consider the limit $\lim_{x\to2}\frac{6x - 5}{3x^2-5x - 2}$. Since $3x^2-5x - 2=(3x + 1)(x - 2)$ and $6x-5=7$ when $x = 2$. The function $y = f(x)$ has a vertical tangent at $x = 2$ and the slope is undefined.

Answer:

$-\frac{15}{2}$

Question 6