Question

to find $f(a)$, we will use the formula $f(a)=lim_{t
ightarrow a}\frac{f(t)-f(a)}{t - a}$. since $f(t)=\frac{3t + 7}{t + 6}$ we have $f(a)=lim_{t
ightarrow a}\frac{\frac{3t + 7}{t + 6}-\frac{3a + 7}{a + 6}}{t - a}$. putting the fractions in the numerator over a common denominator and simplifying, we have $\frac{3t + 7}{t + 6}-\frac{3a + 7}{a + 6}=\frac{(3t + 7)(square)-(3a + 7)(t + 6)}{(t + 6)(a + 6)}=\frac{3ta+7a + 18t + 42-3at - 7t - 18a - 42}{(t + 6)(a + 6)}=\frac{(square)t-(square)a}{(t + 6)(a + 6)}$

Explanation:

Step1: Find common - denominator for the numerator

The common denominator of $\frac{3t + 7}{t + 6}$ and $\frac{3a+7}{a + 6}$ is $(t + 6)(a + 6)$. So $\frac{3t + 7}{t + 6}-\frac{3a + 7}{a + 6}=\frac{(3t + 7)(a + 6)-(3a + 7)(t + 6)}{(t + 6)(a + 6)}$.

Step2: Expand the numerator

Expand $(3t + 7)(a + 6)=3ta+18t + 7a+42$ and $(3a + 7)(t + 6)=3at+18a + 7t+42$. Then the numerator becomes $3ta+18t + 7a+42-(3at + 18a+7t + 42)=3ta+18t + 7a+42 - 3at-18a - 7t - 42=(18 - 7)t-(18 - 7)a = 11t-11a$.

Step3: Find the derivative

We had $f'(a)=\lim_{t
ightarrow a}\frac{\frac{3t + 7}{t + 6}-\frac{3a + 7}{a + 6}}{t - a}=\lim_{t
ightarrow a}\frac{\frac{11t-11a}{(t + 6)(a + 6)}}{t - a}=\lim_{t
ightarrow a}\frac{11(t - a)}{(t - a)(t + 6)(a + 6)}$. Cancel out the $(t - a)$ terms (since $t
eq a$ when taking the limit), we get $f'(a)=\frac{11}{(a + 6)^2}$.

Answer:

$f'(a)=\frac{11}{(a + 6)^2}$