### Calculating [tex]\(\left. \frac{d}{dx} \frac{f(x)}{g(x)+2} \right|_{x=1} \)[/tex]
To find the derivative of [tex]\(\frac{f(x)}{g(x)+2}\)[/tex] at [tex]\(x=1\)[/tex], we'll first express the function and then apply the quotient rule.
The quotient rule states that for two differentiable functions [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex],
[tex]\[
\left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\][/tex]
In our case:
- [tex]\( u(x) = f(x) \)[/tex], so [tex]\( u'(x) = f'(x) \)[/tex]
- [tex]\( v(x) = g(x) + 2 \)[/tex], so [tex]\( v'(x) = g'(x) \)[/tex]
### Step-by-Step Calculation:
1. Evaluate [tex]\(u(x)\)[/tex] and [tex]\(v(x)\)[/tex] at [tex]\(x=1\)[/tex]:
[tex]\[
u(1) = f(1) = -1.8
\][/tex]
[tex]\[
v(1) = g(1) + 2 = 6.9 + 2 = 8.9
\][/tex]
2. Evaluate [tex]\(u'(x)\)[/tex] and [tex]\(v'(x)\)[/tex] at [tex]\(x=1\)[/tex]:
[tex]\[
u'(1) = f'(1) = -1.3
\][/tex]
[tex]\[
v'(1) = g'(1) = -2.3
\][/tex]
3. Apply the quotient rule:
[tex]\[
\left( \frac{f(x)}{g(x) + 2} \right)' \Bigg|_{x=1} = \frac{f'(x)(g(x)+2) - f(x)g'(x)}{(g(x)+2)^2} \Bigg|_{x=1}
\][/tex]
4. Substituting the values:
[tex]\[
\left( \frac{f(x)}{g(x)+2} \right)' \Bigg|_{x=1} = \frac{-1.3 \cdot 8.9 - (-1.8) \cdot (-2.3)}{8.9^2}
\][/tex]
5. Simplify the numerator:
[tex]\[
-1.3 \cdot 8.9 = -11.57
\][/tex]
[tex]\[
-1.8 \cdot -2.3 = 4.14
\][/tex]
[tex]\[
-11.57 - 4.14 = -15.71
\][/tex]
6. Simplify the denominator:
[tex]\[
8.9^2 = 79.21
\][/tex]
7. Combine to find the derivative:
[tex]\[
\left. \frac{d}{dx} \frac{f(x)}{g(x)+2} \right|_{x=1} = \frac{-15.71}{79.21} \approx -0.1983335437444767
\][/tex]
So the derivative [tex]\(\left. \frac{d}{dx} \frac{f(x)}{g(x)+2} \right|_{x=1}\)[/tex] is approximately [tex]\(-0.1983\)[/tex].
