Let's find the derivative of the function [tex]\( g(t) = \frac{1 - 5t}{6 + t} \)[/tex] step by step and match it to the given options.
### Step 1: Identify the Function and Use the Quotient Rule
The function is [tex]\( g(t) = \frac{1 - 5t}{6 + t} \)[/tex]. We will use the quotient rule for differentiation, which is given by:
[tex]\[
\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}
\][/tex]
Here,
- [tex]\( u = 1 - 5t \)[/tex]
- [tex]\( v = 6 + t \)[/tex]
### Step 2: Compute Derivatives of the Numerator and Denominator
First, we compute the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
- Derivative of [tex]\( u = 1 - 5t \)[/tex]:
[tex]\[
u' = \frac{d}{dt}(1 - 5t) = -5
\][/tex]
- Derivative of [tex]\( v = 6 + t \)[/tex]:
[tex]\[
v' = \frac{d}{dt}(6 + t) = 1
\][/tex]
### Step 3: Apply the Quotient Rule
Now, we apply the quotient rule:
[tex]\[
g'(t) = \frac{(1 - 5t)'(6 + t) - (1 - 5t)(6 + t)'}{(6 + t)^2} = \frac{(-5)(6 + t) - (1 - 5t)(1)}{(6 + t)^2}
\][/tex]
### Step 4: Simplify the Expression
Let's simplify the numerator:
[tex]\[
g'(t) = \frac{(-5)(6 + t) - (1 - 5t)}{(6 + t)^2} = \frac{-30 - 5t - 1 + 5t}{(6 + t)^2}
\][/tex]
Notice that [tex]\(-5t\)[/tex] and [tex]\(+5t\)[/tex] cancel each other out:
[tex]\[
g'(t) = \frac{-31}{(6 + t)^2}
\][/tex]
### Step 5: Match with the Given Multiple-Choice Options
We now match our result with the given options:
- a. [tex]\( -\frac{5}{(6 + t)^2} \)[/tex]
- b. [tex]\( -\frac{31}{t^2 + 6} \)[/tex]
- c. [tex]\( -\frac{31}{(6 + t)^2} \)[/tex]
- d. [tex]\( \frac{31}{(6 + t)^2} \)[/tex]
The derivative [tex]\( g'(t) = -\frac{31}{(6 + t)^2} \)[/tex] matches option [tex]\( c \)[/tex].
### Conclusion
Therefore, the correct answer is:
c. [tex]\( -\frac{31}{(6 + t)^2} \)[/tex]
